Differential Equations

$\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x \\ & \text { IF }=e^{2 d x}=e^{2 x} \\ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x+c \\ & =\frac{e^{2 x}}{8}(2 \sin 2 x-2 \cos 2 x)+c \\ & y(0)=\frac{3}{4} \\ & \frac{3}{4}=\frac{1}{8}(-2)+c \Rightarrow c=1 \\ & \text { Put } x=\frac{\pi}{8} \\ & y=\frac{1}{8} \times 2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)+e^{-\pi / 4} \\ & y=e^{-\pi / 4} \end{aligned}$

$\frac{d y}{d x}+\frac{y\left(2 x^3+8 x\right)}{\left(x^2+4\right)^2}=\frac{2}{\left(x^2+4\right)^2}$

$\mathrm{IF}=e^{\int \frac{2 x^3+8 x}{\left(x^2+4\right)^2} d x}$

$\text { Let }\left(x^2+4\right)^2=t \quad \Rightarrow 2\left(x^2+4\right)(2 x) d x=d t$

$\begin{gathered} =e^{\int \frac{d t}{2 t}}=e^{\log \sqrt{t}}=\sqrt{t}=\left(x^2+4\right) \\ \therefore \quad y\left(x^2+4\right)=\int \frac{2}{x^2+4}+c \\ \Rightarrow y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)+c \\ y(0)=0 \end{gathered}$

$\begin{aligned} & \Rightarrow 0=0+c \quad \Rightarrow c=0 \\ & \text { put } x=2 \\ & y(8)=\frac{\pi}{4} \quad \Rightarrow \quad y=\frac{\pi}{32} \\ \end{aligned}$

$\begin{aligned} & \left(x^4+2 x^3+3 x^2+2 x+2\right) d y-\left(2 x^2+2 x+3\right) d x=0 \\ & \int d y=\int\left(\frac{2 x^2+2 x+3}{x^4+2 x^3+3 x^2+2 x+2} d x\right. \\ & \int d y=\int \frac{1}{x^2+1} d x+\int \frac{}{x^2+2 x+2} d x \\ & y=\tan ^{-1}(x)+\tan ^{-1}(1+x)+C \\ & y(-1)=\tan ^{-1}(-1)+\tan ^{-1}(1-1)+C \\ & y(-1)=-\frac{\pi}{4}+C=\left(\frac{-\pi}{4}\right)-\{\text { given }\} \\ & \Rightarrow C=0 \\ & \text { So, } y(x)=\tan ^{-1}(x)+\tan ^{-1}(1+x) \\ & y(0)=\tan ^{-1}(0)+\tan ^{-1}(1+0) \\ & y(0)=\frac{\pi}{4} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\ & \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \\ & {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\ & \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\ & \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\ & \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\ & 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\ & \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\ & \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\ & =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\ & =80+80 \times \frac{1}{8} \\ & =90 \end{aligned}$

$\begin{aligned} & \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\ & =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\ & \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\ & \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\ & \frac{d y}{d x}+p \cdot y=Q \end{aligned}$

$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$

Let $2 \mathrm{x}=\mathrm{t}$

$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$

$\mathrm{dx}=\frac{\mathrm{dt}}{2}$

$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$

$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$

$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$

$\begin{aligned} & y(\text { IF })=\int Q(I F) d x+c \\ & \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\ & y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \\ & y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\ & y=|\tan x|(\ln |\tan x|+c) \\ & \text { Put } x=\frac{\pi}{4}, y=2 \\ & 2=\ln 1+c \Rightarrow c=2 \\ & y=|\tan x|(\ln |\tan x|+2) \\ & y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2) \end{aligned}$

$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}}{\mathrm{y}}\left(\ln \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1\right)$

Let $\frac{x}{y}=t \Rightarrow x=t y$

$\begin{aligned} & \frac{d x}{d y}=t+y \frac{d t}{d y} \\ & t+y \frac{d t}{d y}=t(\ln (t)+1) \end{aligned}$

$\mathrm{y} \frac{\mathrm{dt}}{\mathrm{dy}}=\mathrm{t} \ln (\mathrm{t}) \Rightarrow \frac{\mathrm{dt}}{\mathrm{t} \ln (\mathrm{t})}=\frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t} \cdot \ln (\mathrm{t})}=\int \frac{\mathrm{dy}}{\mathrm{y}}$

$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$ let $\ln t=p$

$\frac{1}{\mathrm{t}} \mathrm{dt}=\mathrm{dp}$

$\begin{aligned} & \Rightarrow \ln p=\ln y+c \\ & \ln (\ln t)=\ln y+c \\ & \ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c \\ & \text { at } x=e, y=1 \\ & \ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0 \end{aligned}$

$\begin{aligned} & \ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y \\ & \left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y} \\ & \left|\ln \left(\frac{x}{y}\right)\right|=y \end{aligned}$

$\frac{d y}{d x}=2(x-1) \sin x+\left(x^2-2 x\right) \cos x$

Now both side integrate

$\begin{aligned} & y(x)=\int 2(x-1) \sin x d x+\left[\left(x^2-2 x\right)(\sin x)-\int(2 x-2) \sin x d x\right] \\ & y(x)=\left(x^2-2 x\right) \sin x+\lambda \\ & y(0)=0+\lambda \Rightarrow 2=\lambda \\ & y(x)=\left(x^2-2 x\right) \sin x+2 \\ & y(2)=2 \end{aligned}$

Differential equation :-

$\begin{aligned} & x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\ & \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x \end{aligned}$

Divide both sides by $\mathrm{x}^2$

$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$

Let $\frac{y}{x}=t$

$\begin{aligned} & \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\ & \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$

Integrating both sides

$\begin{aligned} & \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\ & \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c} \end{aligned}$

Using $\mathrm{y}(1)=\frac{\pi}{3}$, we get $\mathrm{c}=\frac{\sqrt{3}}{2}$

So, $\alpha=\sqrt{3} \Rightarrow \alpha^2=3$

$\begin{aligned} & \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\ & \text { I.F. }=1+\cos ^2 x \\ & y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\ & =-\cos x+C \\ & x=0, C=1 \\ & y\left(\frac{\pi}{2}\right)=1 \end{aligned}$

$\begin{aligned} & \left(x^2-4\right) d y-\left(y^2-3 y\right) d x=0 \\ & \Rightarrow \int \frac{d y}{y^2-3 y}=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3} \int \frac{y-(y-3)}{y(y-3)} d y=\int \frac{d x}{x^2-4} \\ & \Rightarrow \frac{1}{3}(\ln |y-3|-\ln |y|)=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \\ & \Rightarrow \frac{1}{3} \ln \left|\frac{y-3}{y}\right|=\frac{1}{4} \ln \left|\frac{x-2}{x+2}\right|+C \end{aligned}$

$\begin{aligned} & \text { At } \mathrm{x}=4, \mathrm{y}=\frac{3}{2} \\ & \therefore \mathrm{C}=\frac{1}{4} \ln 3 \\ & \therefore \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{\mathrm{x}-2}{\mathrm{x}+2}\right|+\frac{1}{4} \ln (3) \\ & \text { At } \mathrm{x}=10 \\ & \frac{1}{3} \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\frac{1}{4} \ln \left|\frac{2}{3}\right|+\frac{1}{4} \ln (3) \\ & \ln \left|\frac{\mathrm{y}-3}{\mathrm{y}}\right|=\ln 2^{3 / 4}, \forall \mathrm{x}>2, \frac{\mathrm{dy}}{\mathrm{dx}}<0 \\ & \text { as } \mathrm{y}(4)=\frac{3}{2} \Rightarrow \mathrm{y} \in(0,3) \\ & -\mathrm{y}+3=8^{1 / 4} \cdot \mathrm{y} \\ & \mathrm{y}=\frac{3}{1+8^{1 / 4}} \end{aligned}$

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\ & \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\ & \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\ & \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\ & \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\ & \ln 2=0+\mathrm{c} \\ & \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\ & \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\ & \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)} \end{aligned}$

$\begin{aligned} & \frac{d y}{d t}+b y=0 \\ & \frac{d y}{y}=-b d t \\ & \ln |y|=-b t+\lambda \\ & t=0, y=1 \\ & 0=0+\lambda \\ & y=e^{-b t} \quad \text{..... (ii)} \end{aligned}$

According to question

$\begin{aligned} & 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\ & 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\ & \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3} \end{aligned}$

$\begin{aligned} & \text { For } x(t)=y(t) \\ & \begin{array}{r} 2 \mathrm{e}^{-a t}=e^{-b t} \\ 2=e^{(a-b) t} \\ 2=\left(\frac{4}{3}\right)^t \\ \log _{\frac{4}{3}} 2=t \end{array} \end{aligned}$

$ \begin{aligned} & \lim _{1 \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 \\\\ & \Rightarrow \lim _{1 \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1 \\\\ & \Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 \\\\ & \Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9 \\\\ & \Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2} \\\\ & \Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2} \\\\ & \Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x \\\\ & \Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c \\\\ & \because f(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+c \Rightarrow c=\frac{13}{11} \\\\ & \therefore f(x)=\frac{9}{11 x}+\frac{13}{11} x^{10} \end{aligned} $

$\Rightarrow$ Option (B) is correct.

$ \frac{d y}{d x}=\frac{2 x-5 y+3}{5 x+2 y-3} $

Put, $ x=X+h $

and $ y=Y+K $

$ \frac{d y}{d x}=\frac{d Y}{d X}=\frac{2(X+h)-5(Y+K)+3}{5(X+h)+2(Y+K)-3} $

$ \frac{d Y}{d X}=\frac{2 X-5 Y+2 h-5 K+3}{5 X+2 Y+5 h+2 K-3} $

It is Homogeneous differential equation if

$ \begin{aligned} 2 h-5 K+3 & =0 \\\\ 5 h+2 K-3 & =0 \\\\ \frac{h}{15-6} & =\frac{-K}{-6-15}=\frac{1}{4+25} \\\\ \frac{h}{9} & =\frac{K}{21}=\frac{1}{29} \\\\ \Rightarrow \quad h & =\frac{9}{29} \text { and } K=\frac{21}{29} \\\\ \frac{d Y}{d X} & =\frac{2 X-5 Y}{5 X+2 Y} \end{aligned} $

Put $ Y=v X $

$ \begin{array}{c} v+X \frac{d v}{d X}=\frac{2-5 v}{5+2 v} \\\\ \frac{d Y}{d X}=v+X \frac{d v}{d X} \\\\ X \frac{d v}{d X}=\frac{2-5 v}{5+2 v}-v=\frac{2-5 v-5 v-2 v^{2}}{5+2 v} \\\\ X \frac{d v}{d X}=\frac{2-10 v-2 v^{2}}{2 v+5}=-2 \frac{\left(v^{2}+5 v-1\right)}{2 v+5} \\\\ \int \frac{2 v+5}{v^{2}+5 v-1} d v=\int-\frac{2 d x}{x} \\\\ \ln \left(v^{2}+5 v-1\right)=-2 \ln X+\ln C \\\\ \ln \left(v^{2}+5 v-1\right)+\ln X^{2}=\ln C \\\\ x^{2}\left(v^{2}+5 v-1\right)=C \end{array} $

$ \begin{array}{l} X^{2} \frac{\left(Y^{2}+5 X Y-X^{2}\right)}{X^{2}}=C \\\\ Y^{2}+5 X Y-X^{2}=C \\\\ \left(y-\frac{21}{29}\right)^{2}+5\left(x-\frac{9}{29}\right) \\\\ \left(y-\frac{21}{29}\right)-\left(x-\frac{9}{29}\right)^{2}=C \end{array} $

Since, curve passes through point $ (1,1) $

$ \left(1-\frac{21}{29}\right)^{2}+5\left(1-\frac{9}{29}\right) $

$ \left(1-\frac{21}{29}\right)-\left(1-\frac{9}{29}\right)^{2}=C $

$ \left(\frac{8}{29}\right)^{2}+5\left(\frac{20}{29}\right)\left(\frac{8}{29}\right)-\left(\frac{20}{29}\right)^{2}=C $

$ \begin{array}{l} \frac{1}{29^{2}}(64+800-400)=C \\\\ C=\frac{464}{29^2}=\frac{16}{29} \end{array} $

Equation of curve is

$ \begin{array}{l} \left(y-\frac{21}{29}\right)^{2}+5\left(x-\frac{9}{29}\right) \\\\ \left(y-\frac{21}{29}\right)-\left(x-\frac{9}{29}\right)^{2}=\frac{464}{29^{2}} \\\\ y^{2}+\frac{441}{29^{2}}-\frac{42}{29} y +5\left[x y-\frac{9}{29} y-\frac{21}{29} x+\frac{189}{29^{2}}\right] -\left[x^{2}+\frac{81}{29^{2}}-\frac{18 x}{29}\right]=\frac{464}{29^{2}} \end{array} $

$\begin{aligned} & \begin{array}{l}\Rightarrow y^2-x^2+5 x y+x \\\\ \begin{aligned} &\left(-\frac{105}{29}+\frac{18}{29}\right)+y\left(-\frac{42}{29}-\frac{45}{29}\right) +\frac{441+945-81}{29^2}=\frac{464}{29^2}\end{aligned} \\\\ \begin{array}{r}\Rightarrow \quad y^2-x^2+5 x y-\frac{87}{29} x-\frac{87}{29} y +\frac{1305}{29^2}=\frac{464}{29^2}\end{array} \\\\ \begin{array}{c}\Rightarrow y^2-x^2+5 x y-\frac{87}{29} x-\frac{87}{29} y+\frac{841}{29^2}=0\end{array} \\\\ \Rightarrow \quad y^2-x^2+5 x y-\frac{87}{29} x-\frac{87}{29} y+1=0 \\\\ \Rightarrow x^2-y^2-5 x y+\frac{87}{29} x+\frac{87}{29} y-1=0\end{array} \\\\ & x^2-5 x y-y^2+3 x+3 y-1=0\end{aligned}$

$ \left(6 x^{2}-2 x y-18 x+3 y\right) d x $ $ -\left(x^{2}-3 x\right) d y=0 $

$ M(x, y)=6 x^{2}-2 x y-18 x+3 y $

On differentiate $ M $ with respect to $ y $, whe

$ \begin{aligned} \frac{\partial M}{\partial y} & =\frac{d}{d y}\left[6 x^{2}-2 x y-18 x+3 y\right] \\\\ & =-2 x+3 \\\\ N(x, y) & =-\left(x^{2}-3 x\right) \end{aligned} $

On differentiate $ N $ with respect to $ x $, we get

$ \frac{\partial N}{\partial x}=\frac{d}{d x}\left[-\left(x^{2}-3 x\right)\right]=-(2 x-3) $

So, $ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $

The two sites have been shown to be equivalent the equation is an identity any the differential equation is event. Now, let $ \frac{\partial f}{\partial n}=M(x, y) $, then integrate with respect to $ x n $, heat $ y $ as constant as follows

$ \begin{array}{l} \int \frac{\partial f}{\partial x}=\int\left(6 x^{2}-2 x y-18 x+3 y\right) \partial x \\\\ f=2 x^{3}-\frac{y x^{2}}{2}-9 x^{2}+3 x y+H(y) \end{array} $

To solve for $ H(g) $, let $ \frac{\partial f}{\partial y}=N(x, y) $

Then, differentiate of with respect to $ y $, heat $ x $ as constant as follows.

$ \begin{aligned} \frac{\partial f}{\partial y}=0 & -x^{2}-0+3 x+\frac{\partial H(y)}{\partial y} \\\\ & =-x^{2}+3 x+\frac{\partial H(y)}{\partial y} \end{aligned} $

On using Eq. (ii), we get

$ \begin{aligned} -x^{2}+3 x+\frac{\partial H(y)}{\partial y} & =-\left(x^{2}-3 x\right) \\\\ -x^{2}+3 x+\frac{\partial H(y)}{\partial y} & =-x^{2}+3 x \\\\ \frac{\partial H(y)}{\partial y} & =0 \end{aligned} $

Now, integrate the above term as $ \int \partial H(y)=0 $

On substitute $ H(y)=0 $ in the Eq. (i), we get

$ \therefore $ We can equate of to $ C $, we get $ 2 x^{3}-x^{2} y-9 x^{2}+3 x y=C_{1} $

Hence, the general solution of the given differential equation is

$ 2 x^{3}-x^{2} y-9 x^{2}+3 x y+c=0 $

To determine the order and degree of the differential equation, let's start with the given expression:

$ \frac{dy}{dx} = \left(\frac{d^2y}{dx^2} + 2\right)^{\frac{1}{2}} + \frac{d^2y}{dx^2} + 5 $

Rearrange the terms to isolate the square root:

$ \left[\frac{dy}{dx} - \frac{d^2y}{dx^2} - 5\right] = \left(\frac{d^2y}{dx^2} + 2\right)^{\frac{1}{2}} $

Square both sides to eliminate the square root:

$ \left[\frac{dy}{dx} - \frac{d^2y}{dx^2} - 5\right]^2 = \frac{d^2y}{dx^2} + 2 $

Now, we have a polynomial in derivatives. The highest order derivative in this equation is $\frac{d^2y}{dx^2}$, indicating that the order of the differential equation is 2. Since the equation is a polynomial in terms of its highest derivative with the degree being that of the highest exponent applied to the derivative, and here the highest power is 2 (after squaring), the degree of the differential equation is also 2.

Therefore, the order and the degree of the differential equation are both 2.

Given the equation $ y = \sin x + A \cos x $, differentiate both sides with respect to $ x $:

Substitute $ A = \frac{y - \sin x}{\cos x} $ into the equation.

Differentiate:

$ \frac{d y}{d x} = \cos x - \left(\frac{y - \sin x}{\cos x}\right)\sin x $

Simplify:

$ \frac{d y}{d x} = \cos x - y \tan x + \frac{\sin^2 x}{\cos x} $

Further simplification leads to:

$ \frac{d y}{d x} + y \tan x = \frac{\cos^2 x + \sin^2 x}{\cos x} $

Since $\cos^2 x + \sin^2 x = 1$, rewrite it as:

$ \frac{d y}{d x} + y \tan x = \sec x $

Thus, the integrating factor is:

$ e^{\int \tan x \, dx} = e^{\log(\sec x)} = \sec x $

So, the integrating factor of the differential equation is $ \sec x $.

We have,

$ y=A e^{-x}+B \cos x $

On differentiating w.r.t. $ x $, we get

$ y^{\prime}=-A e^{-x}-B \sin x $

On differentiating w.r.t. $ x $, we get

$ y^{\prime \prime}=A e^{-x}-B \cos x $

On adding in Eqs (ii) and (iii), we get

$ \begin{array}{l} y^{\prime \prime}+y^{\prime}=-B(\cos x+\sin x) \\\\ B=\frac{-y^{\prime \prime}-y^{\prime}}{\cos x+\sin x}+x \end{array} $

On adding in Eqs, (i) and (iii), we get

$ \begin{array}{l} y^{\prime \prime}+y=2 A e^{-x} \\\\ e^{-x} A=\frac{y^{\prime \prime}+y}{2} \\\\ \Rightarrow y=\frac{y^{\prime \prime}+y}{2}+\frac{-y^{\prime \prime} \cos x-y^{\prime} \cos x}{\cos x+\sin x} \\\\ \Rightarrow 2 y(\cos x+\sin x)=y^{\prime \prime} \cos x+y^{\prime \prime} \sin x+ \\\\ y \cos x+y \sin x-2 y^{\prime \prime} \cos x-2 y^{\prime} \cos x \\\\ \Rightarrow 2 y(\cos x+\sin x)=y^{\prime \prime}(\sin x-\cos x)+y \\\\ \quad(\cos x+\sin x)-2 y^{\prime} \cos x \\\\ \Rightarrow y^{\prime \prime}(\sin x-\cos x)-y(\cos x+\sin x) \\\\ -2 y^{\prime} \cos x=0 \\\\ \text { Replace } y^{\prime \prime}=\frac{d^{2} y}{d x^{2}}, y^{\prime}=\frac{d y}{d x} \\\\ \because(\sin x-\cos x) \frac{d^{2} y}{d x^{2}}-2 \cos x \frac{d y}{d x}-y \\\\ (\cos x+\sin x)=0 \\\\ \Rightarrow(\cos x-\sin x) \frac{d^{2} y}{d x^{2}}+2 \cos x \frac{d y}{d x}+y \\\\ \quad(\cos x+\sin x)=0 \end{array} $

Given, $ \frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0 $

Let $ 2 x+y=u $

$ \begin{array}{ll} & 2+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-2 \\\\ \Rightarrow & \frac{d u}{d x}-2+\frac{\sin (u)}{\cos x}+2=0 \\\\ \Rightarrow & \frac{d u}{d x}+\frac{\sin u}{\cos x}=0 \end{array} $

By variable seperable

$ \begin{array}{l} \Rightarrow \quad \frac{d u}{\sin u}=-\frac{d x}{\cos x} \\\\ \Rightarrow \quad \int \operatorname{cosec} u d u=-\int \sec x d x \\\\ \Rightarrow \log (\operatorname{cosec} u-\cot u) \\\\ \quad=-\log (\sec x+\tan x)+\log C \\\\ \Rightarrow \log (\operatorname{cosec} u-\cot u) \\\\ \quad+\log (\sec x+\tan x)=\log C \\\\ \Rightarrow(\operatorname{cosec} u-\cot u)(\sec x+\tan x)=C \\\\ \Rightarrow[\operatorname{cosec}(2 x+y)-\cot (2 x+y)] \\\\ \quad[\sec x+\tan x]=C, \end{array} $

$ \begin{aligned} &\text { Given, }\\\\ &\begin{aligned} & (9 x-3 y+5) d y=(3 x-y+1) d x \\\\ & \therefore \quad \frac{d y}{d x}=\frac{3 x-y+1}{9 x-3 y+5} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let } \quad 3 x-y=V\\\\ &\begin{array}{ll} \Rightarrow & 3-\frac{d y}{d x}=\frac{d V}{d x} \\\\ \therefore & 3-\frac{d V}{d x}=\frac{V+1}{3 V+5} \end{array}\\\\ &\begin{array}{lr} \Rightarrow & \frac{d V}{d x}=3-\frac{V+1}{3 V+5}=\frac{8 V+14}{3 V+5} \\\\ \Rightarrow & \left(\frac{3 V+5}{8 V+14}\right) d V=d x \\\\ \Rightarrow & \frac{3}{8}\left(\frac{8 V+\frac{40}{3}}{8 V+14}\right) d V=d x \\\\ \Rightarrow & \frac{3}{8}\left[\frac{8 V+14}{8 V+14}+\frac{\frac{40}{3}-14}{8 V+14}\right] d V=d x \\\\ \Rightarrow & {\left[\frac{3}{8}+\left(-\frac{1}{8}\right)\left(\frac{1}{4 V+7}\right)\right] d V=d x} \end{array} \end{aligned} $

On integrating both sides, we get

$ \begin{aligned} & \frac{3}{8} V-\frac{1}{8} \times \frac{\log |4 V+7|}{4}=x+C_1 \\\\ \Rightarrow \quad & 12(3 x-y)-\log |4(3 x-y)+7| \\\\ \Rightarrow \quad & =32 x+32 C_1 \\\\ \Rightarrow \quad & 4 x-12 y-\log |12 x-4 y+7|=C \end{aligned} $

Where $C=32 C_1$

Given,

$ \begin{aligned} & \frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y} \\\\ \Rightarrow \quad & \frac{d x}{d y}=\frac{2 y^3-4 x y+y}{2 y^2+1} \\\\ \Rightarrow \quad & \frac{d x}{d y}+\left(\frac{4 y}{1+2 y^2}\right) x=\frac{2 y^3+y}{2 y^2+1} \end{aligned} $

This is a linear differential equation

$ \begin{aligned} \text { IF } & =\exp \left(\int \frac{4 y}{1+2 y^2} d y\right) \\\\ & =\exp \left(\log \left(1+2 y^2\right)\right)=1+2 y^2 \end{aligned} $

Now, general solution is given by

$ \begin{aligned} & x \times \mathrm{IF}=\int\left[\left(\frac{2 y^3+y}{2 y^2+1}\right) \times \mathrm{IF}\right] d y+C \\\\ \Rightarrow & x\left(1+2 y^2\right)=\int\left(2 y^3+y\right) d y+C \\\\ \Rightarrow & x+2 x y^2=2\left(\frac{y^4}{4}\right)+\frac{y^2}{2}+C \\\\ \Rightarrow & 4 x y^2+2 x=y^4+y^2+C \end{aligned} $

We have,

$ \begin{aligned} & \left(3 x^2-2 x y\right) d y+\left(y^2-2 x y\right) d x=0 \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{y^2-2 x y}{3 x^2-2 x y}\right) \ldots \text { (i) } \end{aligned} $

Let $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

From Eq. (i), we get

$ \begin{aligned} v+x \frac{d v}{d x} & =-\left(\frac{v^2 x^2-2 v x^2}{3 x^2-2 v x^2}\right) \\\\ x \frac{d v}{d x} & =-\left(\frac{v^2-2 v}{3-2 v}\right)-v \end{aligned} $

$\begin{gathered}x \frac{d v}{d x}=\frac{-v^2+2 v-3 v+2 v^2}{3-2 v} \\\\ x \frac{d v}{d x}=\frac{v^2-v}{3-2 v} \Rightarrow \frac{2 v-3}{v^2-v} d v=-\frac{d x}{x} \\\\ \left.\frac{2 v-1}{v^2-v}-\frac{2}{\left(v-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right) d v=-\frac{d x}{x} \\\\ \log \left(v^2-v\right)-\frac{2}{2\left(\frac{1}{2}\right)} \log \left(\frac{v-\frac{1}{2}-\frac{1}{2}}{v-\frac{1}{2}+\frac{1}{2}}\right)\end{gathered}$

$\begin{aligned} & =-\log x-\log c \\\\ & \log \left(v^2-v\right)-2 \log \left(\frac{v-1}{v}\right)=-(\log c+\log x) \\\\ & \log \left(v^2-v\right)-\log \left(\frac{v-1}{v}\right)^2=-\log c x \\\\ & \Rightarrow \log \left\{\left[v(v-1] \times \frac{v^2}{(v-1)^2}\right\}=-\log \alpha\right.\end{aligned}$

$ \begin{aligned} \frac{v^3}{v-1} & =\frac{1}{c x} \Rightarrow \frac{\frac{y^3}{x^3}}{\frac{y}{x}-1}=\frac{1}{c x} \\\\ \frac{y^3 \cdot x}{x^3(y-x)} & =\frac{1}{c x} \Rightarrow \frac{y^3}{y-x}=\frac{x}{c} \\\\ \Rightarrow \quad x y-x^2 & =c y^3 \end{aligned} $

Which is the general solution of given differential equation.

To determine which option can form a differential equation of order two, consider each geometric configuration:

Option A: All circles passing through the origin

The general equation for a circle passing through the origin is:

$ x^2 + y^2 - 2hx - 2py = 0 \,\,\,\,\,\,\,\,\,\,(i) $

In this equation, there are two arbitrary constants, $ h $ and $ p $. The presence of two arbitrary constants implies that the differential equation derived from this circle's equation would be of order two, since each arbitrary constant contributes to the order of the differential equation.

Conclusion

Among the given options, the circles passing through the origin allow for a differential equation of order two. This is due to the presence of two arbitrary constants in the circle's equation, which dictate the order of the corresponding differential equation.

To find the differential equation that has $ ax + by = 1 $ as its general solution, follow these steps:

Differentiate the Equation: Start by differentiating both sides of $ ax + by = 1 $ with respect to $ x $. This gives:

$ a + b \frac{dy}{dx} = 0 $

Solve for $\frac{dy}{dx}$: From the differentiated equation, solve for $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{-a}{b} $

Differentiate Again: Differentiate the expression $\frac{dy}{dx} = \frac{-a}{b}$ with respect to $ x $:

$ \frac{d^2y}{dx^2} = 0 $

Therefore, the differential equation that has $ ax + by = 1 $ as its general solution is $\frac{d^2y}{dx^2} = 0$.

Given differential equation

$ \begin{array}{ll} e^x y d x+e^x d y+x d x=0 \\ \Rightarrow \left(e^x y+x\right) d x+e^x d y=0 \\ \Rightarrow e^x \frac{d y}{d x}+e^x y+x=0 \\ \Rightarrow \frac{d y}{d x}+y=-\frac{x}{e^x} \end{array} $

which is linear differential equation

$ \therefore \mathrm{IF}=e^{\int P d x}=e^{\int d x}=e^x $

$\therefore$ Required solution

$ \begin{aligned} & y e^x=\int e^x \times\left(\frac{-x}{e^x}\right) d x+C \\ \Rightarrow \quad & y e^x=-\frac{x^2}{2}+C \\ \Rightarrow \quad & 2 y e^x+x^2=C \end{aligned} $

The differential equation for a family of hyperbolas with centers at the origin and axes along the coordinate axes is derived from the standard hyperbola equation:

$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $

Differentiating this equation with respect to $ x $, we get:

$ \frac{2x}{a^2} - \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0 $

This simplifies to:

$ \Rightarrow \frac{2x}{a^2} = \frac{2y}{b^2} \cdot \frac{dy}{dx} \Rightarrow \frac{y}{x} \cdot \frac{dy}{dx} = \frac{b^2}{a^2} $

Differentiating both sides with respect to $ x $ again, we obtain:

$ \frac{y}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^2} = 0 $

Which can be rewritten as:

$ x \cdot y \cdot \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \cdot x - y \cdot \frac{dy}{dx} = 0 $

This results in the differential equation:

$ xyy_2 + y_1^2x - yy_1 = 0 $

To solve the given differential equation:

$ (x y + y^2) \, dx - (x^2 - 2 x y) \, dy = 0 $

we start by rewriting it in the form of a differential equation:

$ \frac{dy}{dx} = \frac{x y + y^2}{x^2 - 2 x y} $

Using the substitution $ y = ux $, where $ u $ is a function of $ x $, we differentiate $ y $ with respect to $ x $:

$ \frac{dy}{dx} = u + x \frac{du}{dx} $

Substituting these into the differential equation provides:

$ u + x \frac{du}{dx} = \frac{u + u^2}{1 - 2u} $

Rearranging terms, we get:

$ x \frac{du}{dx} = \frac{3u^2}{1 - 2u} $

This leads to:

$ \frac{1 - 2u}{3u^2} \, du = \frac{dx}{x} $

Integrating both sides yields:

$ \int \left(\frac{1}{3u^2} - \frac{2}{3u}\right) \, du = \int \frac{dx}{x} $

Calculating the integrals gives:

$ -\frac{1}{3u} - \frac{2}{3} \log u = \log x + \log c $

Substituting back $ u = \frac{y}{x} $:

$ -\frac{x}{3y} - \frac{2}{3} \log \frac{y}{x} = \log xc $

After simplification, the general solution is:

$ c x y^2 e^{\frac{x}{y}} = 1 $

To solve the differential equation $ (1+\tan y)(dx-dy)+2x \, dy = 0 $, we'll start by rewriting the equation:

$ dx + \tan y \, dx - dy - \tan y \, dy + 2x \, dy = 0. $

This simplifies to:

$ -dy(1+\tan y-2x) + dx(1+\tan y) = 0. $

Rewriting the equation, we separate variables:

$ \frac{dy}{dx} = \frac{1+\tan y}{1+\tan y-2x}. $

Alternatively, in terms of $ \frac{dx}{dy} $:

$ \frac{dx}{dy} = 1 - \frac{2x}{1+\tan y}. $

Now, this can be rewritten as:

$ \frac{dx}{dy} + 2 \frac{x}{1+\tan y} = 1. $

This is a first-order linear differential equation of the form:

$ \frac{dx}{dy} + Px = Q $

where $ P = \frac{2}{1+\tan y} $ and $ Q = 1 $.

To solve it, we calculate the integrating factor (IF):

$ \mathrm{IF} = e^{\int P \, dy} = e^{\int \frac{2}{1+\tan y} \, dy}. $

Evaluating the integral, we get:

$ \mathrm{IF} = (\sin y + \cos y) e^y. $

The general solution formula is:

$ x(\mathrm{IF}) = \int Q \cdot (\mathrm{IF}) \, dy + C. $

Substitute the values:

$ x(\sin y+\cos y) e^y = \int 1 \cdot (\sin y+\cos y) e^y \, dy + C. $

Splitting the integral:

$ = \int \sin y \, e^y \, dy + \int \cos y \, e^y \, dy + C. $

Therefore:

$ x(\sin y+\cos y) e^y = \frac{e^y}{2}(\sin y - \cos y) + \frac{e^y}{2}(\cos y + \sin y) + C. $

Simplifying gives:

$ x(\sin y+\cos y) e^y = e^y \cdot \sin y + C. $

This simplifies to:

$ e^y[x \cos y + x \sin y - \sin y] = C. $

Hence, the general solution is:

$ e^y(x \cos y + x \sin y - \sin y) = C. $

$ \begin{aligned} x d y-y d x & =\sqrt{x^2+y^2} d x \\ x d y & =\left(y+\sqrt{x^2+y^2}\right) d x \\ \frac{d y}{d x} & =\frac{y+\sqrt{x^2+y^2}}{x} \end{aligned} $

Let $y=v x$, then $\frac{d y}{d x}=v+x \frac{d v}{d x}$

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{v x+\sqrt{(v x)^2+x^2}}{x} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=v+\sqrt{1+v^2} \\ & \Rightarrow \quad x \frac{d v}{d x}=\sqrt{1+v^2} \\ & \Rightarrow \frac{1}{\sqrt{1+v^2}} \frac{d v}{d x}=\frac{1}{x} \Rightarrow \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log \left|v+\sqrt{1+v^2}\right|=\log x+\log c \\ & \Rightarrow \log \left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log c x \\ & \Rightarrow \log \left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=\log c x \\ & \Rightarrow \log \left|y+\sqrt{\left(x^2+y^2\right)}\right|-\log x=\log c x \\ & \Rightarrow \log \left|y+\sqrt{x^2+y^2}\right|=\log c^2 \\ & \text { or } y+\sqrt{y^2+x^2}=\alpha^2 \end{aligned} $

$x\left(\frac{d^2 y}{d x^2}\right)^{1 / 2}=\left(1+\frac{d y}{d x}\right)^{4 / 3}$

Simplifying, on squaring both sides, we get

$ \begin{aligned} \left(x \frac{d^2 y}{d x^2}\right)^{1 / 2 \times 2} & =\left(1+\frac{d y}{d x}\right)^{4 / 3 \times 2} \\ x^2\left(\frac{d^2 y}{d x^2}\right) & =\left(1+\frac{d y}{d x}\right)^{8 / 3} \end{aligned} $

On cubic both sides, we get

$ \begin{aligned} & \left(x^2 \frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^{8 / 3 \times 3} \\ & x^6\left(\frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^8 \end{aligned} $

Clearly, here the highest derivative with respect to $x$ is 2 , so order is 2 .

Moreover the highest power of highest derivative is 3 i.e. $\left(\frac{d^2 y}{d x^2}\right)^3$, so degree is 3 .

$ \begin{aligned} & \text { Sum }=\text { order }+ \text { degree }=2+3 \\ & \text { Sum }=5 \end{aligned} $

We have,

$ y=A \cos 3 x+B \sin 3 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiate $y$ w.r.t $x$, we get

$ \begin{aligned} & \frac{d y}{d x}=-3 A \sin 3 x+3 B \cos 3 x \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9 A \cos 3 x-9 B \sin 3 x \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9(A \cos 3 x+B \sin 3 x) \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9 y \quad \text { [from Eq. (i)] } \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}+9 y=0 \end{aligned} $

We start with the given differential equation:

$ \cos x \frac{d y}{d x} - y \sin x = 6x, \left(0 < x < \frac{\pi}{2}\right) $

and the initial condition:

$ y\left(\frac{\pi}{3}\right)=0 $

Rewriting the differential equation:

$ \frac{d y}{d x} - y \frac{\sin x}{\cos x} = \frac{6x}{\cos x} $

This simplifies to:

$ \frac{d y}{d x} - y \tan x = 6x \sec x $

This is in the form of the linear differential equation:

$ \frac{d y}{d x} + Ay = B $

where $ A = -\tan x $ and $ B = 6x \sec x $.

The integrating factor (IF) is found using:

$ \mathrm{IF} = e^{\int \operatorname{Adx}} = e^{-\int \tan x \, dx} = e^{-\log \sec x} = \frac{1}{\sec x} $

Applying the integrating factor, the solution becomes:

$ y \cdot \frac{1}{\sec x} = \int 6x \sec x \cdot \frac{1}{\sec x} \, dx $

Simplifying, we get:

$ \frac{y}{\sec x} = 3x^2 + C $

where $ C $ is a constant. Using the initial condition $ y\left(\frac{\pi}{3}\right)=0 $:

$ 0 = 3\left(\frac{\pi}{3}\right)^2 + C \implies C = -\frac{\pi^2}{3} $

Thus, substituting back, we have:

$ \frac{y}{\sec x} = 3x^2 - \frac{\pi^2}{3} $

Solving at $ x = \frac{\pi}{6} $:

$ y = \sec x \cdot \left(3x^2 - \frac{\pi^2}{3}\right) $

Calculating:

$ \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} $

$ 3\left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{12} $

Therefore,

$ y = \frac{2}{\sqrt{3}} \cdot \left(\frac{\pi^2}{12} - \frac{\pi^2}{3}\right) = \frac{2}{\sqrt{3}} \cdot \left(-\frac{\pi^2}{4}\right) $

Simplifying gives:

$ y\left(\frac{\pi}{6}\right) = \frac{-\pi^2}{2\sqrt{3}} $

We have,

$ \frac{d y}{d x}=\frac{y+x \tan \left(\frac{y}{x}\right)}{x} $

Put $y=v x$

$ \begin{array}{ll} \Rightarrow & \frac{d}{d x} \cdot(v x)=\frac{v x+x \tan \left(\frac{v x}{x}\right)}{x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{x(v+\tan v)}{x} \\ \Rightarrow & \frac{x d v}{d x}=v+\tan v-v \\ \Rightarrow & x \frac{d v}{d x}=\tan v \Rightarrow \frac{d v}{\tan v}=\frac{d v}{x} \\ \Rightarrow & \frac{\cos v d v}{\sin v}=\frac{d x}{x} \end{array} $

$ \begin{aligned} &\text { On taking integration both sides, we get }\\ &\begin{aligned} & \Rightarrow \quad \int \frac{\cos v d v}{\sin v}=\int \frac{d x}{x} \\ & \text { Let } \quad \sin v=u \\ & \ \quad \,\,\,\,\,\,\,\,\cos v d v=d u \\ & \Rightarrow \quad \int \frac{1}{u} d u=\int \frac{d x}{x} \\ & \Rightarrow \quad \log u=\log x+\log C \\ & \Rightarrow \quad \log (\sin v)=\log x+\log C \\ & \Rightarrow \quad \log \left(\frac{\sin v}{x}\right)=\log C \Rightarrow \frac{\sin v}{x}=C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & \sin v=x C \\ \Rightarrow & \sin \frac{y}{x}=x C \end{array}\\ &\begin{array}{ll} \Rightarrow & v^2=1.44 \\ \Rightarrow & v=1.2 \mathrm{~m} / \mathrm{s} \end{array} \end{aligned} $

To find the differential equation by eliminating the parameters $ a $ and $ b $ from the equation $ y = a e^{2x} + bx e^{2x} $, we take the derivatives with respect to $ x $ and express them in terms of $ y $.

First Derivative:

$ y' = \frac{d}{dx} (a e^{2x} + bx e^{2x}) = 2a e^{2x} + (2bx e^{2x} + b e^{2x}) $

$ = 2a e^{2x} + 2bx e^{2x} + b e^{2x} = 2y + b e^{2x} $

Second Derivative:

$ y'' = \frac{d}{dx} (2a e^{2x} + 2bx e^{2x} + b e^{2x}) $

$ = 4a e^{2x} + 2b e^{2x} (2x + 1) + b (2 e^{2x}) $

$ = 4a e^{2x} + 4bx e^{2x} + 2b e^{2x} + 2b e^{2x} $

$ = 4a e^{2x} + 4bx e^{2x} + 4b e^{2x} $

$ = 4(a e^{2x} + bx e^{2x}) + 4b e^{2x} $

$ = 4y + 4b e^{2x} $

Constructing the Differential Equation:

Substitute into the expression $ y'' - 4y' + 4y $:

$ y'' - 4y' + 4y = [4y + 4b e^{2x}] - 4[2y + b e^{2x}] + 4y $

$ = 4y + 4b e^{2x} - 8y - 4b e^{2x} + 4y $

$ = 0 $

The expression simplifies to zero, confirming that the differential equation $ y'' - 4y' + 4y = 0 $ is satisfied, supporting the correctness of option (D).

Given the general solution $ y = a^3 e^{b^2 x + c} $, where $ a $ and $ c $ are arbitrary constants and $ b $ is a fixed constant, we are tasked with identifying the order of the differential equation related to this solution.

To find the order, let's derive the expression for $ y $:

$ \begin{aligned} y' &= \frac{d}{dx}\left(a^3 e^{b^2 x + c}\right) \\ &= a^3 b^2 e^{b^2 x + c}. \end{aligned} $

Notice that we can express this in terms of $ y $:

$ y' = b^2 y. $

This differential equation, $ y' - b^2 y = 0 $, is a first-order linear differential equation. The order of a differential equation is determined by the highest derivative present. In this case, since the highest derivative is $ y' $, the order is 1.