3 \(y\left(\frac{d y}{d x}\right)^{2}+y=2 x y \frac{d y}{d x}\)
4 \(y\left(\frac{d y}{d x}\right)^{2}+2 x y \frac{d y}{d x}+y=0\)
Explanation:
(B) : The equation of parabola \(y^{2}=4 a(x+a) \tag{i}\) differentiating w.r.t \(x\) we get- \(2 y \frac{d y}{d x}=4 a \Rightarrow a=\frac{y}{2} \cdot \frac{d y}{d x}\) Putting in equation (i) we get- \(y^{2}=4 \cdot \frac{y}{2} \cdot \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right)\) \(y^{2}=2 y \cdot \frac{d y}{d x} \cdot x+y^{2}\left(\frac{d y}{d x}\right)^{2}\) \(y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}\) \(y\left(\frac{d y}{d x}\right)^{2}+2 x \frac{d y}{d x}-y=0\)
AMU-2021
Differential Equation
87290
If \(\cos \frac{y}{x}=A \log x+C\) is the general solution of \(\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x\), then \(A=\)
1 2
2 1
3 -1
4 -2
Explanation:
(B) : We have given, \(\left(x \sin \frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\) \(\frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin y / x}\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Equation become- \(v+x \frac{d v}{d x}=\frac{v x \sin \frac{v x}{x}-x}{x \sin \frac{v x}{x}} \Rightarrow x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v\) \(x \frac{d v}{d x}=\frac{v \sin v-1-v \sin v}{\sin v} \Rightarrow x \frac{d v}{d x}=\frac{-1}{\sin v}\) On integrating both side we get- \(\int \sin v d x=-\int \frac{d x}{x} \Rightarrow-\cos v=-\log x+c\) Put, \(v=\frac{y}{x}, \cos \left(\frac{y}{x}\right)=\log x+c\) Comparing it by \(\cos (\mathrm{y} / \mathrm{x})=\mathrm{A} \log \mathrm{x}+\mathrm{c}\) \(\mathrm{A}=1\)
AP EAMCET-06.07.2022
Differential Equation
87291
The solution of the equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) satisfying \(y=0\) when \(x=\frac{\pi}{3}\), is
1 \(y=2 \sin ^{2} x+\cos x-2\)
2 \(y=2 \sin ^{2} x-\cos x-2\)
3 \(y=2 \cos ^{2} x-\sin x+2\)
4 \(y=2 \cos x+\sin ^{2} x-1\)
Explanation:
(A) : We have equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is linear equation comparing - \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\mathrm{Q}\) I.F, \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int 2 \tan x} \Rightarrow=\mathrm{e}^{2 \log \sec x} \Rightarrow=\sec ^{2} \mathrm{x}\) For general solution of differential equation, \(y \cdot \sec ^{2} x=\int \sin x \cdot \sec ^{2} x d x+c\) \(y \cdot \sec ^{2} x=\int \frac{\sin x}{\cos x} \cdot \sec x d x+c\) \(y \cdot \sec ^{2} x=\int \tan x \cdot \sec x+c\) \(y \cdot \sec ^{2} x=\sec x+c\) Given, when \(x=\pi / 3\), then \(y=0\) \(0=\sec \pi / 3+c\) \(c+2=0\) \(c=-2\) Now equation become- \(y \sec ^{2} x=\sec x-2\) \(y=\sec x \cdot \frac{1}{\sec ^{2} x}-2 \times \frac{1}{\sec ^{2} x}\) \(y=\cos x-2 \cos ^{2} x\) \(y=\cos x-2\left(1-\sin ^{2} x\right)\) \(y=\cos x-2+2 \sin ^{2} x\) \(y=2 \sin ^{2} x+\cos x-2\)
3 \(y\left(\frac{d y}{d x}\right)^{2}+y=2 x y \frac{d y}{d x}\)
4 \(y\left(\frac{d y}{d x}\right)^{2}+2 x y \frac{d y}{d x}+y=0\)
Explanation:
(B) : The equation of parabola \(y^{2}=4 a(x+a) \tag{i}\) differentiating w.r.t \(x\) we get- \(2 y \frac{d y}{d x}=4 a \Rightarrow a=\frac{y}{2} \cdot \frac{d y}{d x}\) Putting in equation (i) we get- \(y^{2}=4 \cdot \frac{y}{2} \cdot \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right)\) \(y^{2}=2 y \cdot \frac{d y}{d x} \cdot x+y^{2}\left(\frac{d y}{d x}\right)^{2}\) \(y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}\) \(y\left(\frac{d y}{d x}\right)^{2}+2 x \frac{d y}{d x}-y=0\)
AMU-2021
Differential Equation
87290
If \(\cos \frac{y}{x}=A \log x+C\) is the general solution of \(\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x\), then \(A=\)
1 2
2 1
3 -1
4 -2
Explanation:
(B) : We have given, \(\left(x \sin \frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\) \(\frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin y / x}\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Equation become- \(v+x \frac{d v}{d x}=\frac{v x \sin \frac{v x}{x}-x}{x \sin \frac{v x}{x}} \Rightarrow x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v\) \(x \frac{d v}{d x}=\frac{v \sin v-1-v \sin v}{\sin v} \Rightarrow x \frac{d v}{d x}=\frac{-1}{\sin v}\) On integrating both side we get- \(\int \sin v d x=-\int \frac{d x}{x} \Rightarrow-\cos v=-\log x+c\) Put, \(v=\frac{y}{x}, \cos \left(\frac{y}{x}\right)=\log x+c\) Comparing it by \(\cos (\mathrm{y} / \mathrm{x})=\mathrm{A} \log \mathrm{x}+\mathrm{c}\) \(\mathrm{A}=1\)
AP EAMCET-06.07.2022
Differential Equation
87291
The solution of the equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) satisfying \(y=0\) when \(x=\frac{\pi}{3}\), is
1 \(y=2 \sin ^{2} x+\cos x-2\)
2 \(y=2 \sin ^{2} x-\cos x-2\)
3 \(y=2 \cos ^{2} x-\sin x+2\)
4 \(y=2 \cos x+\sin ^{2} x-1\)
Explanation:
(A) : We have equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is linear equation comparing - \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\mathrm{Q}\) I.F, \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int 2 \tan x} \Rightarrow=\mathrm{e}^{2 \log \sec x} \Rightarrow=\sec ^{2} \mathrm{x}\) For general solution of differential equation, \(y \cdot \sec ^{2} x=\int \sin x \cdot \sec ^{2} x d x+c\) \(y \cdot \sec ^{2} x=\int \frac{\sin x}{\cos x} \cdot \sec x d x+c\) \(y \cdot \sec ^{2} x=\int \tan x \cdot \sec x+c\) \(y \cdot \sec ^{2} x=\sec x+c\) Given, when \(x=\pi / 3\), then \(y=0\) \(0=\sec \pi / 3+c\) \(c+2=0\) \(c=-2\) Now equation become- \(y \sec ^{2} x=\sec x-2\) \(y=\sec x \cdot \frac{1}{\sec ^{2} x}-2 \times \frac{1}{\sec ^{2} x}\) \(y=\cos x-2 \cos ^{2} x\) \(y=\cos x-2\left(1-\sin ^{2} x\right)\) \(y=\cos x-2+2 \sin ^{2} x\) \(y=2 \sin ^{2} x+\cos x-2\)
3 \(y\left(\frac{d y}{d x}\right)^{2}+y=2 x y \frac{d y}{d x}\)
4 \(y\left(\frac{d y}{d x}\right)^{2}+2 x y \frac{d y}{d x}+y=0\)
Explanation:
(B) : The equation of parabola \(y^{2}=4 a(x+a) \tag{i}\) differentiating w.r.t \(x\) we get- \(2 y \frac{d y}{d x}=4 a \Rightarrow a=\frac{y}{2} \cdot \frac{d y}{d x}\) Putting in equation (i) we get- \(y^{2}=4 \cdot \frac{y}{2} \cdot \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right)\) \(y^{2}=2 y \cdot \frac{d y}{d x} \cdot x+y^{2}\left(\frac{d y}{d x}\right)^{2}\) \(y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}\) \(y\left(\frac{d y}{d x}\right)^{2}+2 x \frac{d y}{d x}-y=0\)
AMU-2021
Differential Equation
87290
If \(\cos \frac{y}{x}=A \log x+C\) is the general solution of \(\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x\), then \(A=\)
1 2
2 1
3 -1
4 -2
Explanation:
(B) : We have given, \(\left(x \sin \frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\) \(\frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin y / x}\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Equation become- \(v+x \frac{d v}{d x}=\frac{v x \sin \frac{v x}{x}-x}{x \sin \frac{v x}{x}} \Rightarrow x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v\) \(x \frac{d v}{d x}=\frac{v \sin v-1-v \sin v}{\sin v} \Rightarrow x \frac{d v}{d x}=\frac{-1}{\sin v}\) On integrating both side we get- \(\int \sin v d x=-\int \frac{d x}{x} \Rightarrow-\cos v=-\log x+c\) Put, \(v=\frac{y}{x}, \cos \left(\frac{y}{x}\right)=\log x+c\) Comparing it by \(\cos (\mathrm{y} / \mathrm{x})=\mathrm{A} \log \mathrm{x}+\mathrm{c}\) \(\mathrm{A}=1\)
AP EAMCET-06.07.2022
Differential Equation
87291
The solution of the equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) satisfying \(y=0\) when \(x=\frac{\pi}{3}\), is
1 \(y=2 \sin ^{2} x+\cos x-2\)
2 \(y=2 \sin ^{2} x-\cos x-2\)
3 \(y=2 \cos ^{2} x-\sin x+2\)
4 \(y=2 \cos x+\sin ^{2} x-1\)
Explanation:
(A) : We have equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is linear equation comparing - \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\mathrm{Q}\) I.F, \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int 2 \tan x} \Rightarrow=\mathrm{e}^{2 \log \sec x} \Rightarrow=\sec ^{2} \mathrm{x}\) For general solution of differential equation, \(y \cdot \sec ^{2} x=\int \sin x \cdot \sec ^{2} x d x+c\) \(y \cdot \sec ^{2} x=\int \frac{\sin x}{\cos x} \cdot \sec x d x+c\) \(y \cdot \sec ^{2} x=\int \tan x \cdot \sec x+c\) \(y \cdot \sec ^{2} x=\sec x+c\) Given, when \(x=\pi / 3\), then \(y=0\) \(0=\sec \pi / 3+c\) \(c+2=0\) \(c=-2\) Now equation become- \(y \sec ^{2} x=\sec x-2\) \(y=\sec x \cdot \frac{1}{\sec ^{2} x}-2 \times \frac{1}{\sec ^{2} x}\) \(y=\cos x-2 \cos ^{2} x\) \(y=\cos x-2\left(1-\sin ^{2} x\right)\) \(y=\cos x-2+2 \sin ^{2} x\) \(y=2 \sin ^{2} x+\cos x-2\)
3 \(y\left(\frac{d y}{d x}\right)^{2}+y=2 x y \frac{d y}{d x}\)
4 \(y\left(\frac{d y}{d x}\right)^{2}+2 x y \frac{d y}{d x}+y=0\)
Explanation:
(B) : The equation of parabola \(y^{2}=4 a(x+a) \tag{i}\) differentiating w.r.t \(x\) we get- \(2 y \frac{d y}{d x}=4 a \Rightarrow a=\frac{y}{2} \cdot \frac{d y}{d x}\) Putting in equation (i) we get- \(y^{2}=4 \cdot \frac{y}{2} \cdot \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right)\) \(y^{2}=2 y \cdot \frac{d y}{d x} \cdot x+y^{2}\left(\frac{d y}{d x}\right)^{2}\) \(y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}\) \(y\left(\frac{d y}{d x}\right)^{2}+2 x \frac{d y}{d x}-y=0\)
AMU-2021
Differential Equation
87290
If \(\cos \frac{y}{x}=A \log x+C\) is the general solution of \(\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x\), then \(A=\)
1 2
2 1
3 -1
4 -2
Explanation:
(B) : We have given, \(\left(x \sin \frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\) \(\frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin y / x}\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Equation become- \(v+x \frac{d v}{d x}=\frac{v x \sin \frac{v x}{x}-x}{x \sin \frac{v x}{x}} \Rightarrow x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v\) \(x \frac{d v}{d x}=\frac{v \sin v-1-v \sin v}{\sin v} \Rightarrow x \frac{d v}{d x}=\frac{-1}{\sin v}\) On integrating both side we get- \(\int \sin v d x=-\int \frac{d x}{x} \Rightarrow-\cos v=-\log x+c\) Put, \(v=\frac{y}{x}, \cos \left(\frac{y}{x}\right)=\log x+c\) Comparing it by \(\cos (\mathrm{y} / \mathrm{x})=\mathrm{A} \log \mathrm{x}+\mathrm{c}\) \(\mathrm{A}=1\)
AP EAMCET-06.07.2022
Differential Equation
87291
The solution of the equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) satisfying \(y=0\) when \(x=\frac{\pi}{3}\), is
1 \(y=2 \sin ^{2} x+\cos x-2\)
2 \(y=2 \sin ^{2} x-\cos x-2\)
3 \(y=2 \cos ^{2} x-\sin x+2\)
4 \(y=2 \cos x+\sin ^{2} x-1\)
Explanation:
(A) : We have equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is linear equation comparing - \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\mathrm{Q}\) I.F, \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int 2 \tan x} \Rightarrow=\mathrm{e}^{2 \log \sec x} \Rightarrow=\sec ^{2} \mathrm{x}\) For general solution of differential equation, \(y \cdot \sec ^{2} x=\int \sin x \cdot \sec ^{2} x d x+c\) \(y \cdot \sec ^{2} x=\int \frac{\sin x}{\cos x} \cdot \sec x d x+c\) \(y \cdot \sec ^{2} x=\int \tan x \cdot \sec x+c\) \(y \cdot \sec ^{2} x=\sec x+c\) Given, when \(x=\pi / 3\), then \(y=0\) \(0=\sec \pi / 3+c\) \(c+2=0\) \(c=-2\) Now equation become- \(y \sec ^{2} x=\sec x-2\) \(y=\sec x \cdot \frac{1}{\sec ^{2} x}-2 \times \frac{1}{\sec ^{2} x}\) \(y=\cos x-2 \cos ^{2} x\) \(y=\cos x-2\left(1-\sin ^{2} x\right)\) \(y=\cos x-2+2 \sin ^{2} x\) \(y=2 \sin ^{2} x+\cos x-2\)
3 \(y\left(\frac{d y}{d x}\right)^{2}+y=2 x y \frac{d y}{d x}\)
4 \(y\left(\frac{d y}{d x}\right)^{2}+2 x y \frac{d y}{d x}+y=0\)
Explanation:
(B) : The equation of parabola \(y^{2}=4 a(x+a) \tag{i}\) differentiating w.r.t \(x\) we get- \(2 y \frac{d y}{d x}=4 a \Rightarrow a=\frac{y}{2} \cdot \frac{d y}{d x}\) Putting in equation (i) we get- \(y^{2}=4 \cdot \frac{y}{2} \cdot \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right)\) \(y^{2}=2 y \cdot \frac{d y}{d x} \cdot x+y^{2}\left(\frac{d y}{d x}\right)^{2}\) \(y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^{2}\) \(y\left(\frac{d y}{d x}\right)^{2}+2 x \frac{d y}{d x}-y=0\)
AMU-2021
Differential Equation
87290
If \(\cos \frac{y}{x}=A \log x+C\) is the general solution of \(\left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x\), then \(A=\)
1 2
2 1
3 -1
4 -2
Explanation:
(B) : We have given, \(\left(x \sin \frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x\) \(\frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin y / x}\) Let, \(\mathrm{y}=\mathrm{vx}\) \(\frac{d y}{d x}=v+x \frac{d v}{d x}\) Equation become- \(v+x \frac{d v}{d x}=\frac{v x \sin \frac{v x}{x}-x}{x \sin \frac{v x}{x}} \Rightarrow x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v\) \(x \frac{d v}{d x}=\frac{v \sin v-1-v \sin v}{\sin v} \Rightarrow x \frac{d v}{d x}=\frac{-1}{\sin v}\) On integrating both side we get- \(\int \sin v d x=-\int \frac{d x}{x} \Rightarrow-\cos v=-\log x+c\) Put, \(v=\frac{y}{x}, \cos \left(\frac{y}{x}\right)=\log x+c\) Comparing it by \(\cos (\mathrm{y} / \mathrm{x})=\mathrm{A} \log \mathrm{x}+\mathrm{c}\) \(\mathrm{A}=1\)
AP EAMCET-06.07.2022
Differential Equation
87291
The solution of the equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) satisfying \(y=0\) when \(x=\frac{\pi}{3}\), is
1 \(y=2 \sin ^{2} x+\cos x-2\)
2 \(y=2 \sin ^{2} x-\cos x-2\)
3 \(y=2 \cos ^{2} x-\sin x+2\)
4 \(y=2 \cos x+\sin ^{2} x-1\)
Explanation:
(A) : We have equation \(\frac{d y}{d x}+2 y \tan x=\sin x\) Which is linear equation comparing - \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{py}=\mathrm{Q}\) I.F, \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int 2 \tan x} \Rightarrow=\mathrm{e}^{2 \log \sec x} \Rightarrow=\sec ^{2} \mathrm{x}\) For general solution of differential equation, \(y \cdot \sec ^{2} x=\int \sin x \cdot \sec ^{2} x d x+c\) \(y \cdot \sec ^{2} x=\int \frac{\sin x}{\cos x} \cdot \sec x d x+c\) \(y \cdot \sec ^{2} x=\int \tan x \cdot \sec x+c\) \(y \cdot \sec ^{2} x=\sec x+c\) Given, when \(x=\pi / 3\), then \(y=0\) \(0=\sec \pi / 3+c\) \(c+2=0\) \(c=-2\) Now equation become- \(y \sec ^{2} x=\sec x-2\) \(y=\sec x \cdot \frac{1}{\sec ^{2} x}-2 \times \frac{1}{\sec ^{2} x}\) \(y=\cos x-2 \cos ^{2} x\) \(y=\cos x-2\left(1-\sin ^{2} x\right)\) \(y=\cos x-2+2 \sin ^{2} x\) \(y=2 \sin ^{2} x+\cos x-2\)