Chapter 24

If for \(x \geq 0, y=y(x)\) is the solution of the differential equation, \((x+1) d y=\left((x+1)^{2}+y-3\right) d x, y(2)=0\) then \(y(3)\) is equal to

Let \(y=y(x)\) be the solution curve of the differential equation, \(\left(y^{2}-x\right) \frac{d y}{d x}=1\), satisfying \(y(0)=1\). This curve intersects the \(x\)-axis at a point whose abscissa is: (a) \(2-e\) (b) \(-e\) (c) 2 (d) \(2+e\)

Consider the differential equation, \(y^{2} d x+\left(x-\frac{1}{y}\right) d y=0\). If value of \(y\) is 1 when \(x=1\), then the value of \(x\) for which \(y=2\), is : (a) \(\frac{5}{2}+\frac{1}{\sqrt{e}}\) (b) \(\frac{3}{2}-\frac{1}{\sqrt{e}}\) (c) \(\frac{1}{2}+\frac{1}{\sqrt{e}}\) (d) \(\frac{3}{2}-\sqrt{e}\)

If \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) is the solution of the differential equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=(\tan x-y) \sec ^{2} x, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(\mathrm{y}(0)=0\) then \(y\left(-\frac{\pi}{4}\right)\) is equal to: (a) \(\mathrm{e}-2\) (b) \(\frac{1}{2}-e\) (c) \(2+\frac{1}{e}\) (d) \(\frac{1}{e}-2 \mid\)

Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(y(0)\) \(=1\). Then: (a) \(y\left(\frac{\pi}{4}\right)+y\left(-\frac{\pi}{4}\right)=\frac{\pi^{2}}{2}+2\) (b) \(y^{\prime}\left(\frac{\pi}{4}\right)+y^{\prime}\left(-\frac{\pi}{4}\right)=-\sqrt{2}\) (c) \(y\left(\frac{\pi}{4}\right)-y\left(-\frac{\pi}{4}\right)=\sqrt{2}\) (d) \(y^{\prime}\left(\frac{\pi}{4}\right)-y^{\prime}\left(-\frac{\pi}{4}\right)=\pi-\sqrt{2}\)

The solution of the differential equation \(x \frac{d y}{d x}+2 y=x^{2}\) \((x \neq 0)\) with \(y(1)=1\), is: (a) \(y=\frac{4}{5} x^{3}+\frac{1}{5 x^{2}}\) (b) \(y=\frac{x^{3}}{5}+\frac{1}{5 x^{2}}\) (c) \(y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}\) (d) \(y=\frac{3}{4} x^{2}+\frac{1}{4 x^{2}}\)

Let \(y=y(x)\) be the solution of the differential equation, \(\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{a}\) \(y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is: (a) \(\frac{1}{4}\) (b) \(\frac{1}{2}\) (c) 1 (d) \(\frac{1}{16}\)

Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be the solution of the differential equation, \(x \frac{d y}{d x}+y=x \log _{e} \mathrm{x},(x>1) .\) If \(2 y(2)=\log _{e} 4-1\), then \(y(e)\) is equal to: [Jan. 12, 2019 (I)] (a) \(-\frac{e}{2}\) (b) \(-\frac{e^{2}}{2}\) (c) \(\frac{e}{4}\) (d) \(\frac{e^{2}}{4}\)

If a curve passes through the point \((1,-2)\) and has slope of the tangent at any point \((x, y)\) on it as \(\frac{x^{2}-2 y}{x}\), then the curve also passes through the point : (a) \((3,0)\) (b) \((\sqrt{3}, 0)\) (c) \((-1,2)\) (d) \((-\sqrt{2}, 1)\)

If \(y(x)\) is the solution of the differential equation \(\frac{d y}{d x}+\left(\frac{2 x+1}{x}\right) y=e^{-2 x}, x>0\), where \(y(1)=\frac{1}{2} e^{-2}\), then (a) \(y\left(\log _{e} 2\right)=\log _{e} 4\) (b) \(y\left(\log _{e} 2\right)=\frac{\log _{e} 2}{4}\) (c) \(y(x)\) is decreasing in \(\left(\frac{1}{2}, 1\right)\) (d) \(\mathrm{y}(x)\) is decreasing in \((0,1)\)

Let \(f\) be a differentiable function such that \(f^{\prime}(x)=7-\frac{3}{4} \frac{f(x)}{x}\), \((x>0)\) and \(f(1) \neq 4\). Then \(\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)\) (a) exists and equals \(\frac{4}{7}\). (b) exists and equals 4 . (c) does not exist. (d) exists and equals 0 .

If \(y=y(x)\) is the solution of the differential equation, \(x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=x^{2}\) satisfying \(y(1)=1\), then \(y\left(\frac{1}{2}\right)\) is equal to: (a) \(\frac{7}{64}\) (b) \(\frac{1}{4}\) (c) \(\frac{49}{16}\) (d) \(\frac{13}{16}\)

Let \(y-y(x)\) be the solution of the differential equation \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to: (a) \(\frac{-8}{9 \sqrt{3}} \pi^{2}\) (b) \(-\frac{8}{9} \pi^{2}\) (c) \(-\frac{4}{9} \pi^{2}\) (d) \(\frac{4}{9 \sqrt{3}} \pi^{2}\)

Let \(y=y(x)\) be the solution of the differential equation \(\frac{d y}{d x}+2 y=f(x)\), where \(f(x)= \begin{cases}1, & x \in[0,1] \\ 0, & \text { otherwise }\end{cases}\) If \(y(0)=0\), then \(y\left(\frac{3}{2}\right)\) is (a) \(\frac{e^{2}-1}{2 e^{3}}\) (b) \(\frac{e^{2}-1}{e^{3}}\) (c) \(\frac{1}{2 e}\) (d) \(\frac{e^{2}+1}{2 e^{4}}\)

The curve satisfying the differential equation, \(\mathrm{ydx}-\mathrm{x}+\) 3y \(^{2}\) ) \(d y=0\) and passing through the point \((1,1)\), also passes through the point: (a) \(\left(\frac{1}{4},-\frac{1}{2}\right)\) (b) \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) (c) \(\left(\frac{1}{3},-\frac{1}{3}\right)\) (d) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)

The solution of the differential equaiton \(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{2} \sec \mathrm{x}=\frac{\tan \mathrm{x}}{2 \mathrm{y}}\), where \(0 \leq \mathrm{x}<\frac{\pi}{2}\), and \(\mathrm{y}(0)=1\), is given by: (a) \(y^{2}=1+\frac{x}{\sec x+\tan x}\) (b) \(y=1+\frac{x}{\sec x+\tan x}\) (c) \(y=1-\frac{x}{\sec x+\tan x}\) (d) \(y^{2}=1-\frac{x}{\sec x+\tan x}\)

Let \(y(x)\) be the solution of the differential equation \((x \log x) \frac{d y}{d x}+y=2 x \log x,(x \geq 1)\). Then \(y(e)\) is equal to: (a) 2 (d) 0 (b) \(2 \mathrm{e}\) (c) \(\mathrm{e}\)

If \(\frac{d y}{d x}+y \tan x=\sin 2 x\) and \(y(0)=1\), then \(y(\pi)\) is equal to: (a) 1 (c) \(-5\) (b) \(-1\) (d) 5

The general solution of the differential equation, \(\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0\) (a) \(y \sqrt{\tan x}=x+c\) (b) \(y \sqrt{\cot x}=\tan x+c\) (c) \(y \sqrt{\tan x}=\cot x+c\) (d) \(y \sqrt{\cot x}=x+c\)

The equation of the curve passing through the origin and satisfying the differential equation \(\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2}\) is (a) \(\left(1+x^{2}\right) y=x^{3}\) (b) \(3\left(1+x^{2}\right) y=2 x^{3}\) (c) \(\left(1+x^{2}\right) y=3 x^{3}\) (d) \(3\left(1+x^{2}\right) y=4 x^{3}\)

The integrating factor of the differential equation \(\left(x^{2}-1\right) \frac{d y}{d x}+2 x y=x\) is (a) \(\frac{1}{x^{2}-1}\) (b) \(x^{2}-1\) (c) \(\frac{x^{2}-1}{x}\) (d) \(\frac{x}{x^{2}-1}\)

Consider the differential equation \(y^{2} d x+\left(x-\frac{1}{y}\right) d y=0 .\) If \(y(1)=1\), then \(x\) is given by: (a) \(4-\frac{2}{y}-\frac{e^{\frac{1}{y}}}{e}\) (b) \(3-\frac{1}{y}+\frac{e^{\frac{1}{y}}}{e}\) (c) \(1+\frac{1}{y}-\frac{e^{\frac{1}{y}}}{e}\) (d) \(1-\frac{1}{y}+\frac{e^{\frac{1}{y}}}{e}\)

Solution of the differential equation \(\cos x d y=y(\sin x-y) d x, 0

Solution of the differential equation \(y d x+\left(x+x^{2} y\right) d y=0\) is (a) \(\log y=C x\) (b) \(-\frac{1}{x y}+\log y=C\) (c) \(\frac{1}{x y}+\log y=C\) (d) \(-\frac{1}{x y}=C\)

The solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\), is (a) \(x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+k\) (b) \((x-2)=k e^{2 \tan ^{-1} y}\) (c) \(2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k\) (d) \(x e^{\tan ^{-1} y}=\tan ^{-1} y+k\)