Differential Equations
Let the solution curve $y=f(x)$ of the differential equation $ \frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}$, $x\in(-1,1)$ pass through the origin. Then $\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x $ is equal to
If ${{dy} \over {dx}} + 2y\tan x = \sin x,\,0 < x < {\pi \over 2}$ and $y\left( {{\pi \over 3}} \right) = 0$, then the maximum value of $y(x)$ is :
Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1,2)$ and $(8,1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to
The slope of the tangent to a curve $C: y=y(x)$ at any point $(x, y)$ on it is $\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$. If $C$ passes through the points $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$ and $\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$, then $\mathrm{e}^{\alpha}$ is equal to :
The general solution of the differential equation $\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$ is :
Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$, represents a circle with center ($\alpha$, $\beta$). Then, $\alpha$ + 2$\beta$ is equal to :
If y = y(x) is the solution of the differential equation $\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$ and y (0) = 0, then $6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$ is equal to
Let the solution curve of the differential equation
$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}} $, $y(1) = 3$ be $y = y(x)$. Then y(2) is equal to:
Let x = x(y) be the solution of the differential equation
$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$ such that x(1) = 0. Then, x(e) is equal to :
Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $\tan x(\cos x - y)$. If the curve passes through the point $\left( {{\pi \over 4},0} \right)$, then the value of $\int\limits_0^{\pi /2} {y\,dx} $ is equal to :
Let the solution curve $y = y(x)$ of the differential equation
$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$
pass through the points (1, 0) and (2$\alpha$, $\alpha$), $\alpha$ > 0. Then $\alpha$ is equal to
Let y = y(x) be the solution of the differential equation $x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$, $x > 1$, with $y(2) = - 2$. Then y(3) is equal to :
If the solution curve of the differential equation
$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is
Let ${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :
If ${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$, x, y > 0, y(1) = 1, then y(2) is equal to :
If $y = y(x)$ is the solution of the differential equation
$x{{dy} \over {dx}} + 2y = x\,{e^x}$, $y(1) = 0$ then the local maximum value
of the function $z(x) = {x^2}y(x) - {e^x},\,x \in R$ is :
If the solution of the differential equation
${{dy} \over {dx}} + {e^x}\left( {{x^2} - 2} \right)y = \left( {{x^2} - 2x} \right)\left( {{x^2} - 2} \right){e^{2x}}$ satisfies $y(0) = 0$, then the value of y(2) is _______________.
If $y = y(x)$ is the solution of the differential equation
$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$ such that $y(e) = {e \over 3}$, then y(1) is equal to :
Let $g:(0,\infty ) \to R$ be a differentiable function such that
$\int {\left( {{{x(\cos x - \sin x)} \over {{e^x} + 1}} + {{g(x)\left( {{e^x} + 1 - x{e^x}} \right)} \over {{{({e^x} + 1)}^2}}}} \right)dx = {{x\,g(x)} \over {{e^x} + 1}} + c} $, for all x > 0, where c is an arbitrary constant. Then :
Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$, with $y(0) = {1 \over 3}$. Then, the point $x = - {4 \over 3}$ for the curve $y = y(x)$ is :
If the solution curve $y = y(x)$ of the differential equation ${y^2}dx + ({x^2} - xy + {y^2})dy = 0$, which passes through the point (1, 1) and intersects the line $y = \sqrt 3 x$ at the point $(\alpha ,\sqrt 3 \alpha )$, then value of ${\log _e}(\sqrt 3 \alpha )$ is equal to :
If x = x(y) is the solution of the differential equation
$y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0$; then x(e) is equal to :
Let $y=y(x)$ be the solution curve of the differential equation
$\sin \left( {2{x^2}} \right){\log _e}\left( {\tan {x^2}} \right)dy + \left( {4xy - 4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \right)dx = 0$, $0 < x < \sqrt {{\pi \over 2}} $, which passes through the point $\left(\sqrt{\frac{\pi}{6}}, 1\right)$. Then $\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|$ is equal to ______________.
Explanation:
${{dy} \over {dx}} + y\left( {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}} \right) = {{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})\ln (\tan {x^2})}}$
$I.F = {e^{\int {{{4x} \over {\sin (2{x^2})\ln (\tan {x^2})}}dx} }}$
$ = {e^{\ln |\ln (\tan {x^2})}} = \ln (\tan {x^2})$
$\therefore$ $\int {d(y.\ln (\tan {x^2})) = \int {{{4\sqrt 2 x\sin \left( {{x^2} - {\pi \over 4}} \right)} \over {\sin (2{x^2})}}dx} } $
$ \Rightarrow y\ln (\tan {x^2}) = \ln \left| {{{\sec {x^2} + \tan {x^2}} \over {{\mathop{\rm cosec}\nolimits} \,{x^2} - \cot {x^2}}}} \right| + C$
$\ln \left( {{1 \over {\sqrt 3 }}} \right) = \ln \left( {{{{3 \over {\sqrt 3 }}} \over {2 - \sqrt 3 }}} \right) + C$
$e = \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$
For $y\left( {\sqrt {{\pi \over 3}} } \right)$
$y\ln \left( {\sqrt 3 } \right) = \ln \left| {{{2 + \sqrt 3 } \over {{2 \over {\sqrt 3 }} + {1 \over {\sqrt 3 }}}}} \right| + \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2\sqrt 3 }}} \right)$
$ = \ln \left( {2 + \sqrt 3 } \right) + \ln \left( {{1 \over {\sqrt 3 }}} \right) + \ln \left( {{1 \over {\sqrt 3 }}} \right) - \ln \left( {{{\sqrt 3 } \over {2 - \sqrt 3 }}} \right)$
$ \Rightarrow y\ln \sqrt 3 = \ln \left( {{1 \over {\sqrt 3 }}} \right)$
$ \Rightarrow {y \over 2}\ln 3 = - {1 \over 2}\ln 3$
$ \Rightarrow y = 1$
$\therefore$ $\left| {y\left( {\sqrt {{\pi \over 3}} } \right)} \right| = 1$.
Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{d y}{d x}-y=2-e^{-x}$ such that $\lim\limits_{x \rightarrow \infty} y(x)$ is finite. If $a$ and $b$ are respectively the $x$ - and $y$-intercepts of the tangent to the curve at $x=0$, then the value of $a-4 b$ is equal to _____________.
Explanation:
IF $ = {e^{-x}}$
$y\,.\,{e^{-x}} = - 2{e^{ - x}} + {{{e^{ - 2x}}} \over 2} + C$
$ \Rightarrow y = - 2 + {e^{ - x}} + C{e^x}$
$\mathop {\lim }\limits_{x \to \infty } \,y(x)$ is finite so $C = 0$
$y = - 2 + {e^{ - x}}$
$ \Rightarrow {\left. {{{dy} \over {dx}} = - {e^{ - x}} \Rightarrow {{dy} \over {dx}}} \right|_{x = 0}} = - 1$
Equation of tangent
$y + 1 = - 1(x - 0)$
or $y + x = - 1$
So $a = - 1,\,b = - 1$
$ \Rightarrow a - 4b = 3$
Let a curve $y=y(x)$ pass through the point $(3,3)$ and the area of the region under this curve, above the $x$-axis and between the abscissae 3 and $x(>3)$ be $\left(\frac{y}{x}\right)^{3}$. If this curve also passes through the point $(\alpha, 6 \sqrt{10})$ in the first quadrant, then $\alpha$ is equal to ___________.
Explanation:
$\int\limits_3^x {f(x)dx = {{\left( {{{f(x)} \over x}} \right)}^3}} $
${x^3}\,.\,\int\limits_3^x {f(x)dx = {f^3}(x)} $
Differentiate w.r.t. x
${x^3}f(x) + 3{x^2}\,.\,{{{f^3}(x)} \over {{x^3}}} = 3{f^2}(x)f'(x)$
$ \Rightarrow 3{y^2}{{dy} \over {dx}} = {x^3}y + {{3{y^3}} \over x}$
$3xy{{dy} \over {dx}} = {x^4} + 3{y^2}$
Let ${y^2} = t$
${3 \over 2}{{dt} \over {dx}} = {x^3} + {{3t} \over x}$
${{dt} \over {dx}} - {{2t} \over x} = {{2{x^3}} \over 3}$
$I.F. = \,.\,{e^{\int { - {2 \over x}dx} }} = {1 \over {{x^2}}}$
Solution of differential equation
$t\,.\,{1 \over {{x^2}}} = \int {{2 \over 3}x\,dx} $
${{{y^2}} \over {{x^2}}} = {{{x^2}} \over 3} + C$
${y^2} = {{{x^4}} \over 3} + C{x^2}$
Curve passes through $(3,3) \Rightarrow C = - 2$
${y^2} = {{{x^4}} \over 3} - 2{x^2}$
Which passes through $\left( {\alpha ,6\sqrt {10} } \right)$
${{{\alpha ^4} - 6{\alpha ^2}} \over 3} = 360$
${\alpha ^4} - 6{\alpha ^2} - 1080 = 0$
$\alpha = 6$
Let $y=y(x)$ be the solution of the differential equation
$\frac{d y}{d x}=\frac{4 y^{3}+2 y x^{2}}{3 x y^{2}+x^{3}}, y(1)=1$.
If for some $n \in \mathbb{N}, y(2) \in[n-1, n)$, then $n$ is equal to _____________.
Explanation:
${{dy} \over {dx}} = {y \over x}{{(4{y^2} + 2{x^2})} \over {(3{y^2} + {x^2})}}$
Put $y = vx$
$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$
$ \Rightarrow v + x{{dv} \over {dx}} = {{v(4{v^2} + 2)} \over {(3{v^2} + 1)}}$
$ \Rightarrow x{{dv} \over {dx}} = v\left( {{{(4{v^2} + 2 - 3{v^2} - 1)} \over {3{v^2} + 1}}} \right)$
$ \Rightarrow \int {(3{v^2} + 1){{dv} \over {{v^3} + v}} = \int {{{dx} \over x}} } $
$ \Rightarrow \ln |{v^3} + v| = \ln x + c$
$ \Rightarrow \ln \left| {{{\left( {{y \over x}} \right)}^3} + \left( {{y \over x}} \right)} \right| = \ln x + C$
$ \downarrow \,y(1) = 1$
$ \Rightarrow C = \ln 2$
$\therefore$ for $y(2)$
$\ln \left( {{{{y^3}} \over 8} + {y \over 2}} \right) = 2\ln 2 \Rightarrow {{{y^3}} \over 8} + {y \over 2} = 4$
$ \Rightarrow [y(2)] = 2$
$ \Rightarrow n = 3$
Let y = y(x), x > 1, be the solution of the differential equation $(x - 1){{dy} \over {dx}} + 2xy = {1 \over {x - 1}}$, with $y(2) = {{1 + {e^4}} \over {2{e^4}}}$. If $y(3) = {{{e^\alpha } + 1} \over {\beta {e^\alpha }}}$, then the value of $\alpha + \beta $ is equal to _________.
Explanation:
$ \text { I.F} =e^{\int \frac{2 x}{x-1} d x} $
$ \begin{aligned} & =e^{2 \int\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right) d x} \\\\ & =e^{2 x+2 \ln (x-1)} \\\\ & =\mathrm{e}^{2 x}(\mathrm{x}-1)^{2} \end{aligned} $
$\Rightarrow \int d\left(y \cdot e^{2 x}(x-1)^{2}\right)=\int e^{2 x} d x$
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}}{2}+c$
$\downarrow y(2)=\frac{1+e^{4}}{2 e^{4}}$
$\frac{1+e^{4}}{2 e^{4}} \cdot e^{4}=\frac{e^{4}}{2}+c$
$\Rightarrow \quad c=\frac{e^{4}}{2}\left(\frac{1+e^{4}-e^{4}}{e^{4}}\right)=\frac{1}{2}$
$\Rightarrow \quad y \cdot e^{2 x}(x-1)^{2}=\frac{e^{2 x}+1}{2}$
$ \downarrow y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}} $
$\Rightarrow \frac{e^{\alpha}+1}{\beta e^{\alpha}} \cdot e^{6} \cdot 4=\frac{e^{6}+1}{2}$
$\Rightarrow \alpha=6$ and $\beta=8 \Rightarrow \alpha+\beta=14$
Let y = y(x) be the solution of the differential equation ${{dy} \over {dx}} + {{\sqrt 2 y} \over {2{{\cos }^4}x - {{\cos }^2}x}} = x{e^{{{\tan }^{ - 1}}(\sqrt 2 \cot 2x)}},\,0 < x < {\pi \over 2}$ with $y\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {32}}$. If $y\left( {{\pi \over 3}} \right) = {{{\pi ^2}} \over {18}}{e^{ - {{\tan }^{ - 1}}(\alpha )}}$, then the value of 3$\alpha$2 is equal to ___________.
Explanation:
I.F. $=e^{\int \frac{2 \sqrt{2} d x}{1+\cos ^{2} 2 x}}=e^{\sqrt{2} \int \frac{2 \sec ^{2} 2 x}{2+\tan ^{2} 2 x} d x}$ $=e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}$
$\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=\int x e^{\tan ^{-1}(\sqrt{2} \cot 2 x)}$
$ \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)} d x+c $
$\Rightarrow y \cdot e^{\tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)}=e^{\frac{\pi}{2}} \cdot \frac{x^{2}}{2}+c$
When $x=\frac{\pi}{4}, y=\frac{\pi^{2}}{32}$ gives $c=0$
When $x=\frac{\pi}{3}, y=\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha}$
So $\frac{\pi^{2}}{18} e^{-\tan ^{-1} \alpha} \cdot e^{-\tan ^{-1}\left(-\sqrt{\left.\frac{3}{2}\right)}\right.}=e^{\pi / 2} \frac{\pi^{2}}{18}$
$\Rightarrow \quad-\tan ^{-1} \alpha+\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}(-\alpha)=\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)$
$\Rightarrow \alpha=-\sqrt{\frac{2}{3}} \Rightarrow 3 \alpha^{2}=2$
Let $y = y(x)$ be the solution of the differential equation $(1 - {x^2})dy = \left( {xy + ({x^3} + 2)\sqrt {1 - {x^2}} } \right)dx, - 1 < x < 1$, and $y(0) = 0$. If $\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = k} $, then k$-$1 is equal to _____________.
Explanation:
$\left( {1 - {x^2}} \right)dy = \left( {xy + \left( {{x^3} + 2} \right)\sqrt {1 - {x^2}} } \right)dx$
$\therefore$ ${{dy} \over {dx}} - {x \over {1 - {x^2}}}y = {{{x^3} + 3} \over {\sqrt {1 - {x^2}} }}$
$\therefore$ $I.F. = {e^{\int { - {x \over {1 - {x^2}}}dx} }} = \sqrt {1 - {x^2}} $
Solution is
$y.\,\sqrt {1 - {x^2}} = \int {\left( {{x^3} + 3} \right)dx} $
$y.\,\sqrt {1 - {x^2}} = {{{x^4}} \over 4} + 3x + c$
$\because$ $y(0) = 0 \Rightarrow c = 0$
$\therefore$ $y(x) = {{{x^4} + 12x} \over {4\sqrt {1 - {x^2}} }}$
$\therefore$ $\int_{{{ - 1} \over 2}}^{{1 \over 2}} {\sqrt {1 - {x^2}} y(x)dx = \int_{{{ - 1} \over 2}}^{{1 \over 2}} {\left( {{{{x^4} + 12x} \over 4}} \right)dx = \int_0^{{1 \over 2}} {{{{x^4}} \over 2}dx} } } $
$\therefore$ $k = {1 \over {320}}$
$\therefore$ $ = {k^{ - 1}} = 320$
Let the solution curve y = y(x) of the differential equation
$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$ pass through the origin. Then y(2) is equal to _____________.
Explanation:
$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$
$ \Rightarrow {{dy} \over {dx}} = \left( {{{6x} \over {{x^2} + 4}}} \right)y + 2x$
$ \Rightarrow {{dy} \over {dx}} - \left( {{{6x} \over {{x^2} + 4}}} \right)y = 2x$
$I.F. = {e^{ - 3\ln ({x^2} + 4)}} = {1 \over {{{({x^2} + 4)}^3}}}$
So ${y \over {{{({x^2} + 4)}^3}}} = \int {{{2x} \over {{{({x^2} + 4)}^3}}}dx + c} $
$ \Rightarrow y = - {1 \over 2}({x^2} + 4) + c{({x^2} + 4)^3}$
When x = 0, y = 0 gives $c = {1 \over {32}}$,
So, for x = 2, y = 12
Let $S = (0,2\pi ) - \left\{ {{\pi \over 2},{{3\pi } \over 4},{{3\pi } \over 2},{{7\pi } \over 4}} \right\}$. Let $y = y(x)$, x $\in$ S, be the solution curve of the differential equation ${{dy} \over {dx}} = {1 \over {1 + \sin 2x}},\,y\left( {{\pi \over 4}} \right) = {1 \over 2}$. If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve $y = \sqrt 2 \sin x$ is ${{k\pi } \over {12}}$, then k is equal to _____________.
Explanation:
${{dy} \over {dx}} = {1 \over {1 + \sin 2x}}$
$ \Rightarrow dy = {{{{\sec }^2}xdx} \over {{{(1 + \tan x)}^2}}}$
$ \Rightarrow y = - {1 \over {1 + \tan x}} + c$
When $x = {\pi \over 4}$, $y = {1 \over 2}$ gives c = 1
So $y = {{\tan x} \over {1 + \tan x}} \Rightarrow y = {{\sin x} \over {\sin x + \cos x}}$
Now, $y = \sqrt 2 \sin x \Rightarrow \sin x = 0$
or $\sin x + \cos x = {1 \over {\sqrt 2 }}$
$\sin x = 0$ gives $x = \pi $ only.
and $\sin x + \cos x = {1 \over {\sqrt 2 }} \Rightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over 2}$
So $x + {\pi \over 4} = {{5\pi } \over 6}$ or ${{13\pi } \over 6} \Rightarrow x = {{7\pi } \over {12}}$ or ${{23\pi } \over {12}}$
Sum of all solutions $ = \pi + {{7\pi } \over {12}} + {{23\pi } \over {12}} = {{42\pi } \over {12}}$
Hence, $k = 42$.
${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$ such that y(0) = 7. Then y($\pi$) is equal to :
2x2 dy + (ey $-$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
equation (x $-$ x3)dy = (y + yx2 $-$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$, with y(0) = 0.
If $y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$, then the value of $\alpha$ is equal to :
equation xdy = (y + x3 cosx)dx with y($\pi$) = 0, then $y\left( {{\pi \over 2}} \right)$ is equal to :
then, the minimum value of $y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$ is equal to :
${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$, 0 < x < 2.1, with y(2) = 0. Then the value of ${{dy} \over {dx}}$ at x = 1 is equal to :