Differential Equations

We have $\frac{d y}{d x}+x \sin 2 y=x^3 \cos ^2 y$

Dividing both sides by $\cos ^2 y$, we get

$ \begin{aligned} & \sec ^2 y \frac{d y}{d x}+x \frac{\sin 2 y}{\cos ^2 y}=x^3 \\ & \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3 \end{aligned} $

Put $\tan y=v$

and $\sec ^2 y \frac{d y}{d x}=\frac{d v}{d x}$, we get

$ \frac{d v}{d x}+(2 x) v=x^3 $

Now, $\mathrm{IF}=e^{\int 2 x d x}=e^{x^2}$

Multiplying both sides by IF, we get

$ e^{x^2} \frac{d v}{d x}+2 x v \cdot e^{x^2}=e^{x^2} \cdot x^3 $

$ \begin{aligned} &\text { Integrating both sides w.r.t } x \text {, we get }\\ &\begin{aligned} & v e^{x^2}=\int x^3 e^{x^2} d x+C \\ &=\int x^3 e^t\left(\frac{d t}{2 x}\right)+C, \text { where } t=x^2 \\ &=\frac{1}{2} \int t e^t d t+C=\frac{1}{2}\left(t e^t-e^t\right)+C \\ &=\frac{1}{2}(t-1) e^t+C=\frac{1}{2}\left(x^2-1\right) e^{x^2}+C \\ & \Rightarrow \quad e^{x^2} \tan y=\frac{1}{2}\left(x^2-1\right) e^{x^2}+C \end{aligned} \end{aligned} $

To solve the given differential equation

$ (x+1) \frac{d y}{d x} - y = e^{3x}(x+1)^2 $

we start by rewriting it in the standard linear form:

$ \frac{d y}{d x} - \frac{y}{x+1} = e^{3x}(x+1) $

Here, it's a first-order linear differential equation of the form

$ \frac{d y}{d x} + P(x)y = Q(x) $

where $ P(x) = -\frac{1}{x+1} $ and $ Q(x) = e^{3x}(x+1) $.

Step 1: Calculate the Integrating Factor (IF)

The integrating factor is given by:

$ \text{IF} = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x+1} \, dx} = e^{-\ln(x+1)} = \frac{1}{x+1} $

Step 2: Determine the Solution

Using the integrating factor, the solution of the differential equation is:

$ y \cdot \text{IF} = \int Q(x) \cdot \text{IF} \, dx $

Substituting the known values:

$ y \cdot \frac{1}{x+1} = \int e^{3x}(x+1) \cdot \frac{1}{x+1} \, dx $

This simplifies to:

$ \frac{y}{x+1} = \frac{e^{3x}}{3} + C' $

Therefore, the complete solution is:

$ \frac{3y}{x+1} = e^{3x} + C' $

where $ C' = C/3 $ representing an integration constant.

Given, $(1+x)^2 d y+2 x y d x=\cot x d x$

$\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{\cot x}{1+x^2}$, which is a linear differential equation of the form of $\frac{d y}{d x}+P y=Q$

Here, $P=\frac{2 x}{1+x^2}$ and $Q=\frac{\cot x}{1+x^2}$

Now, IF $=\exp \left(\int \frac{2 x}{1+x^2} d x\right)=\exp \left(\log \left(1+x^2\right)\right)$

$=1+x^2$

and the solution of this linear differential equation be

$\begin{aligned} & y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C \\ & \Rightarrow \quad y\left(1+x^2\right)=\int\left(\frac{\cot x}{1+x^2} \times\left(1+x^2\right)\right) d x+C \\ & \Rightarrow \quad y\left(1+x^2\right)=\int \cot x d x+C \\ & \Rightarrow \quad y\left(1+x^2\right)=\log |\sin x|+C \\ & \Rightarrow \quad y=\frac{\log |\sin x|}{1+x^2}+\frac{C}{1+x^2} \end{aligned}$

Given that

$\left( {{{dy} \over {dx}}} \right)\tan x = y{\sec ^2}x + \sin x$

$\tan \left( {{{dy} \over {dx}}} \right) - y{\sec ^2}x = \sin x$

${{dy} \over {dx}} + \left( {{{ - {{\sec }^2}x} \over {\tan x}}} \right)y = {{\sin x} \over {\tan x}}$

Here, $P = - {{{{\sec }^2}x} \over {\tan x}}$ and $Q = \cos x$

$IF = {e^{\int {P\,dx} }} = {e^{\int {{{ - {{\sec }^2}x} \over {\tan x}}dx} }}$

$IF = {e^{ - \log (\tan x)}} = {e^{\log \cot x}} = \cot x$

Now, $y \times IF = \int {(Q \times IF)dx + C} $

$y\,.\,\cot x = \int {\cos x\,.\,\cot x + C} $

$y\,.\,\cot x = \int {{{1 - {{\sin }^2}x} \over {\sin x}} + C} $

$y\,.\,\cot x = \log |{\mathop{\rm cosec}\nolimits} \,x - \cot x| + \cos x + C$

$\left( {{{x + y - 1} \over {x + y - 2}}} \right){{dy} \over {dx}} = \left( {{{x + y + 1} \over {x + y + 2}}} \right)$

Put $x + y = v$

$\therefore$ $1 + {{dy} \over {dx}} = {{dv} \over {dx}}$ or $\left( {{{v - 1} \over {v - 2}}} \right)\left( {{{dv} \over {dx}} - 1} \right) = \left( {{{v + 1} \over {v + 2}}} \right)$

$ \Rightarrow {{dv} \over {dx}} - 1 = {{(v + 1)(v - 2)} \over {(v - 1)(v + 2)}} = {{{v^2} - v - 2} \over {{v^2} + v - 2}}$ or ${{dv} \over {dx}} = {{2{v^2} - 4} \over {({v^2} + v - 2)}}$

$ \Rightarrow {{({v^2} + v - 2)} \over {({v^2} - 2)}}dv = 2dx \Rightarrow \left( {1 + {v \over {{v^2} - 2}}} \right)dv = 2dx$

On integrating, we get

$v + {1 \over 2}\log \left| {{v^2} - 2} \right| = 2x + c$

Given, y = 1 when x = 1

$\therefore$ $0 + {1 \over 2}\log 2 = c$ or $(y - x) + {1 \over 2}\log \left| {{{{{(x + y)}^2} - 2} \over 2}} \right| = 0$

or, $\log \left| {{{{{(x + y)}^2} - 2} \over 2}} \right| = 2(x - y)$

We have, $\cos y{{dy} \over {dx}} + {4 \over x}\sin y = {{{e^x}} \over {{x^3}}}$

Let $\sin y = t$

$ \Rightarrow \cos y{{dy} \over {dx}} = {{dt} \over {dx}}$ $\therefore$ ${{dt} \over {dx}} + {4 \over x}t = {{{e^x}} \over {{x^3}}}$

$IF = {e^{\int {{4 \over x}dx} }} = {e^{4\log x}} = {x^4}$

The solution is

$\therefore$ $t{x^4} = \int {{x^4}\,.\,{{{e^x}} \over {{x^3}}}dx = x{e^x} - {e^x} + c} $ $\because$ $\sin y\,{x^4} = x{e^x} - {e^x} + c$

$\because$ x = 1, y = 0 [$\because$ c = 0]

$\therefore$ $\sin y = {e^x}(x - 1){x^{ - 4}}$

We have, ${{dy} \over {dx}} + {{\tan y} \over x} = {{\tan y\sin y} \over {{x^2}}}$

Now, divide by tan y sin y both sides, we get

$\cot y\cos ec\,y{{dy} \over {dx}} + {{\cos ec\,y} \over x} = {1 \over {{x^2}}}$ ...... (i)

Put cosec y = z

$\therefore$ $ - \cos ec\,y\cot y{{dy} \over {dx}} = {{dz} \over {dx}}$

$\therefore$ Eq. (i) reduces to

${{dz} \over {dx}} - {z \over x} = - {1 \over {{x^2}}}$

$IF = {e^{\int { - {1 \over x}dx} }} = {e^{ - \log x}} = {1 \over x}$

Hence solution is

${z \over x} = \int { - {1 \over {{x^3}}}dx = {1 \over {2{x^2}}} + c} $

$ \Rightarrow {1 \over {x\sin y}} = {{1 - 2c{x^2}} \over {2{x^2}}}$

$ \Rightarrow 2x = \sin y{(1 - 2cx)^2}$

$ \Rightarrow 2x = \sin y(1 - 2c{x^2})$

We have, $(x{y^5} + 2y)dx = x\,dy$

$ \Rightarrow x{{dy} \over {dx}} - 2y = x{y^5}$

$ \Rightarrow {{dy} \over {dx}} - {{2y} \over x} = {y^5}$

$ \Rightarrow {y^{ - 5}}{{dy} \over {dx}} - {{2{y^{ - 4}}} \over x} = 1$ ..... (i)

Put, ${y^{ - 4}} = t$

$ \Rightarrow - 4{y^{ - 5}}{{dy} \over {dx}} = {{dt} \over {dx}}$

$ \Rightarrow {y^{ - 5}}{{dy} \over {dx}} = {{ - 1} \over 4}{{dt} \over {dx}}$ ..... (ii)

From Eqs. (i) and (ii), we get

$ - {1 \over 4}{{dt} \over {dx}} - {{2t} \over x} = 1$

$ \Rightarrow {{dt} \over {dx}} + {{8t} \over x} = - 4$

Now, $IF = {e^{\int {{8 \over x}dx} }} = {e^{8\log x}} = {x^8}$

and the solution is

$t\,.\,{x^8} = \int {( - 4){x^8}dx + C} $

$ \Rightarrow {{{x^8}} \over {{y^4}}} = - {{4\,.\,{x^9}} \over 9} + C$

$ \Rightarrow 9{x^8} + 4{x^9}\,.\,{y^4} = 9{y^4}C$

We have,

Slope of tangent of the curve is ${{{{(x + 1)}^2} + y - 3} \over {x + 1}}$

$\therefore$ ${{dy} \over {dx}} = {{{{(x + 1)}^2} + y - 3} \over {x + 1}}$

$ \Rightarrow {{dy} \over {dx}} = {{{{(x + 1)}^2} - 3} \over {x + 1}} + {y \over {x + 1}}$

$ \Rightarrow {{dy} \over {dx}} - {y \over {x + 1}} = {{{{(x + 1)}^2} - 3} \over {x + 1}}$

This is a form of linear differential equations

$\therefore$ $IF = {e^{\int {{{ - 1} \over {x + 1}}dx} }} = {e^{\log {1 \over {1 + x}}}} = {1 \over {1 + x}}$

Solution of given differential equation is

${y \over {x + 1}} = \int {{{{{(x + 1)}^2} - 3} \over {{{(x + 1)}^2}}}dx} $

${y \over {x + 1}} = \int {\left( {1 - {3 \over {{{(x + 1)}^2}}}} \right)dx} $

${y \over {x + 1}} = x + {3 \over {x + 1}} + c$

$\because$ Curve passes through (2, 0)

$\therefore$ $0 = 2 + {3 \over 3} + c$

$ \Rightarrow c = - 3$

Hence, equation of curve is

${y \over {x + 1}} = x + {3 \over {x + 1}} - 3$

$y = {x^2} + x + 3 - 3x - 3$

$y = {x^2} - 2x$