87270
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}=\frac{y}{x}\) \(\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3\). Then \(\frac{y^{2}(x)}{9}\) is equal to:
87284
\(y+x^{2}=\frac{d y}{d x}\) has the solution
1 \(y+x^{2}+2 x+2=c e^{x}\)
2 \(y+x+2 x^{2}+2=c e^{x}\)
3 \(y^{2}+x+x^{2}+2=c e^{2 x}\)
4 \(y+x+x^{2}+2=c e^{2 x}\)
Explanation:
(A) : We have- \(y+x^{2}=d y / d x\) It can be written as \(\mathrm{dy} / \mathrm{dx}-\mathrm{y}=\mathrm{x}^{2}\) Which is a linear differential equation. Where, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{x}^{2}\) \(I \cdot F=e^{\int p d x}=e^{\int(-1) d x} \Rightarrow I \cdot F=e^{-x}\) Now for general solution- \(y \times I F=\int I F \times Q d x+c \Rightarrow y \cdot e^{-x}=\int x^{2} \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+\int 2 x \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+2\left[-\mathrm{xe}^{-x}-\left(e^{-x}\right)\right]+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+c\) \(\text { dividing by e } e^{-x} \text { then we get- }\) \(y=-x^{2}-2 x-2+e^{x}\)
AP EAMCET-2002
Differential Equation
87285
The general solution of \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\)
1 \(y-x^{2}=c \sec x\)
2 \(y \cos x=x^{2} \sec x+c\)
3 \(y \sec x=x^{2}+c \cos x\)
4 \(y=x^{2}+c \cos x\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\) Which is a linear differential equation \(P=\tan x \text { and } Q=2 x+x^{2} \tan x\) I.F \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int \tan \mathrm{x}} \Rightarrow=\mathrm{e}^{\log \sec x}=\sec \mathrm{x}\) For solution of equation - \(y \cdot \sec x=\int \sec x\left(2 x+x^{2} \tan x\right) d x+c\) \(y \cdot \sec x=\int\left(2 x \sec x+x^{2} \sec x \tan x\right) d x+c\) \(y \cdot \sec x=\int \frac{d}{d x}\left(x^{2} \sec x\right) d x+c\) \(y \sec x=x^{2} \sec x+c\) \(y=x^{2}+c \cos x\)
AP EAMCET-22.04.2018
Differential Equation
87271
The solution of the differential equation \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0\) is
1 \(\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0\)
2 \(\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0\)
3 \(\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0\)
4 \(\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
Explanation:
(D) : We have differential equation- \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right)\) Let \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) Then equation, \(v+x \frac{d v}{d x}=-\left(\frac{x^{2}+3 v^{2} x^{2}}{3 x^{2}+v^{2} x^{2}}\right)\) \(v+x \frac{d v}{d x}=-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=-v-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=\frac{-3 v-v^{3}-1-3 v^{2}}{3+v^{2}}\) \(x \frac{d v}{d x}=-\frac{(v+1)^{3}}{3+v^{2}}\) Separating the variable we get- \(\frac{3+v^{2}}{(v+1)^{3}} d v=-\frac{d x}{x}\) On integrating both side we get- \(\int \frac{3+v^{2}}{(v+1)^{3}} d v=-\int \frac{d x}{x}\) \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{A}{(v+1)}+\frac{B}{(v+1)^{2}}+\frac{C}{(v+1)^{3}}\) \(3+v^{2}=A\left(v^{2}+1+2 v\right)+B(v+1)+C\) \(A=1\) \(2 A+B=0, B=-2\) \(A+B+C=3\) \(1+(-2)+C=3\) \(\quad C=4\) Then, \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{1}{v+1}-\frac{2}{(v+1)^{2}}+\frac{4}{(v+1)^{3}}\) The equation becomes- \(-\int \frac{d x}{x}=\frac{1}{v+1} d v-\int \frac{2}{(v+1)^{2}} d v+\int \frac{4}{(v+1)^{3}} d v\) \(-\log x+\log c=\log (v+1)+\frac{2}{v+1}-\frac{4}{2(v+1)^{2}}\) Now substituting, \(v=y / x\) \(\log \mathrm{c} / \mathrm{x}=\log \left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)+\frac{2}{\frac{\mathrm{y}}{\mathrm{x}}+1}-\frac{2}{\left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)^{2}}\) \(\log \mathrm{c} / \mathrm{x}-\log \left(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{x}}\right)=\frac{2 \mathrm{x}^{2}+2 \mathrm{xy}-2 \mathrm{x}^{2}}{(\mathrm{x}+\mathrm{y})^{2}}\) \(\log \left(\frac{c}{x+y}\right)=\frac{2 x y}{(x+y)^{2}}\) \(-\log \left|\frac{x+y}{c}\right|=\frac{2 x y}{(x+y)^{2}}\) \(\log |x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
JEE Main-30.01.2023
Differential Equation
87272
The general solution of the differential equation \(\left(1+y^{2}\right) \mathbf{d x}+\left(1+\mathbf{x}^{2}\right) \mathbf{d y}=\mathbf{0}\) is
1 \(x-y=c(1-x y)\)
2 \(x-y=c(1-x y)\)
3 \(x+y=c(1-x y)\)
4 \(x+y=c(1+x y)\)
Explanation:
(C) : Differential equation \(\left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0\) On dividing of \(\left(1+x^{2}\right)\left(1+y^{2}\right)\) in above equation \(\frac{d x}{1+x^{2}}+\frac{d y}{1+y^{2}}=0\) On integrating both side, we get \(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}\) \(\frac{x+y}{1-x y}=c\) \(x+y=c(1-x y)\)
87270
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}=\frac{y}{x}\) \(\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3\). Then \(\frac{y^{2}(x)}{9}\) is equal to:
87284
\(y+x^{2}=\frac{d y}{d x}\) has the solution
1 \(y+x^{2}+2 x+2=c e^{x}\)
2 \(y+x+2 x^{2}+2=c e^{x}\)
3 \(y^{2}+x+x^{2}+2=c e^{2 x}\)
4 \(y+x+x^{2}+2=c e^{2 x}\)
Explanation:
(A) : We have- \(y+x^{2}=d y / d x\) It can be written as \(\mathrm{dy} / \mathrm{dx}-\mathrm{y}=\mathrm{x}^{2}\) Which is a linear differential equation. Where, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{x}^{2}\) \(I \cdot F=e^{\int p d x}=e^{\int(-1) d x} \Rightarrow I \cdot F=e^{-x}\) Now for general solution- \(y \times I F=\int I F \times Q d x+c \Rightarrow y \cdot e^{-x}=\int x^{2} \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+\int 2 x \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+2\left[-\mathrm{xe}^{-x}-\left(e^{-x}\right)\right]+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+c\) \(\text { dividing by e } e^{-x} \text { then we get- }\) \(y=-x^{2}-2 x-2+e^{x}\)
AP EAMCET-2002
Differential Equation
87285
The general solution of \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\)
1 \(y-x^{2}=c \sec x\)
2 \(y \cos x=x^{2} \sec x+c\)
3 \(y \sec x=x^{2}+c \cos x\)
4 \(y=x^{2}+c \cos x\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\) Which is a linear differential equation \(P=\tan x \text { and } Q=2 x+x^{2} \tan x\) I.F \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int \tan \mathrm{x}} \Rightarrow=\mathrm{e}^{\log \sec x}=\sec \mathrm{x}\) For solution of equation - \(y \cdot \sec x=\int \sec x\left(2 x+x^{2} \tan x\right) d x+c\) \(y \cdot \sec x=\int\left(2 x \sec x+x^{2} \sec x \tan x\right) d x+c\) \(y \cdot \sec x=\int \frac{d}{d x}\left(x^{2} \sec x\right) d x+c\) \(y \sec x=x^{2} \sec x+c\) \(y=x^{2}+c \cos x\)
AP EAMCET-22.04.2018
Differential Equation
87271
The solution of the differential equation \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0\) is
1 \(\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0\)
2 \(\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0\)
3 \(\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0\)
4 \(\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
Explanation:
(D) : We have differential equation- \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right)\) Let \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) Then equation, \(v+x \frac{d v}{d x}=-\left(\frac{x^{2}+3 v^{2} x^{2}}{3 x^{2}+v^{2} x^{2}}\right)\) \(v+x \frac{d v}{d x}=-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=-v-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=\frac{-3 v-v^{3}-1-3 v^{2}}{3+v^{2}}\) \(x \frac{d v}{d x}=-\frac{(v+1)^{3}}{3+v^{2}}\) Separating the variable we get- \(\frac{3+v^{2}}{(v+1)^{3}} d v=-\frac{d x}{x}\) On integrating both side we get- \(\int \frac{3+v^{2}}{(v+1)^{3}} d v=-\int \frac{d x}{x}\) \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{A}{(v+1)}+\frac{B}{(v+1)^{2}}+\frac{C}{(v+1)^{3}}\) \(3+v^{2}=A\left(v^{2}+1+2 v\right)+B(v+1)+C\) \(A=1\) \(2 A+B=0, B=-2\) \(A+B+C=3\) \(1+(-2)+C=3\) \(\quad C=4\) Then, \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{1}{v+1}-\frac{2}{(v+1)^{2}}+\frac{4}{(v+1)^{3}}\) The equation becomes- \(-\int \frac{d x}{x}=\frac{1}{v+1} d v-\int \frac{2}{(v+1)^{2}} d v+\int \frac{4}{(v+1)^{3}} d v\) \(-\log x+\log c=\log (v+1)+\frac{2}{v+1}-\frac{4}{2(v+1)^{2}}\) Now substituting, \(v=y / x\) \(\log \mathrm{c} / \mathrm{x}=\log \left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)+\frac{2}{\frac{\mathrm{y}}{\mathrm{x}}+1}-\frac{2}{\left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)^{2}}\) \(\log \mathrm{c} / \mathrm{x}-\log \left(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{x}}\right)=\frac{2 \mathrm{x}^{2}+2 \mathrm{xy}-2 \mathrm{x}^{2}}{(\mathrm{x}+\mathrm{y})^{2}}\) \(\log \left(\frac{c}{x+y}\right)=\frac{2 x y}{(x+y)^{2}}\) \(-\log \left|\frac{x+y}{c}\right|=\frac{2 x y}{(x+y)^{2}}\) \(\log |x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
JEE Main-30.01.2023
Differential Equation
87272
The general solution of the differential equation \(\left(1+y^{2}\right) \mathbf{d x}+\left(1+\mathbf{x}^{2}\right) \mathbf{d y}=\mathbf{0}\) is
1 \(x-y=c(1-x y)\)
2 \(x-y=c(1-x y)\)
3 \(x+y=c(1-x y)\)
4 \(x+y=c(1+x y)\)
Explanation:
(C) : Differential equation \(\left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0\) On dividing of \(\left(1+x^{2}\right)\left(1+y^{2}\right)\) in above equation \(\frac{d x}{1+x^{2}}+\frac{d y}{1+y^{2}}=0\) On integrating both side, we get \(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}\) \(\frac{x+y}{1-x y}=c\) \(x+y=c(1-x y)\)
87270
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}=\frac{y}{x}\) \(\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3\). Then \(\frac{y^{2}(x)}{9}\) is equal to:
87284
\(y+x^{2}=\frac{d y}{d x}\) has the solution
1 \(y+x^{2}+2 x+2=c e^{x}\)
2 \(y+x+2 x^{2}+2=c e^{x}\)
3 \(y^{2}+x+x^{2}+2=c e^{2 x}\)
4 \(y+x+x^{2}+2=c e^{2 x}\)
Explanation:
(A) : We have- \(y+x^{2}=d y / d x\) It can be written as \(\mathrm{dy} / \mathrm{dx}-\mathrm{y}=\mathrm{x}^{2}\) Which is a linear differential equation. Where, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{x}^{2}\) \(I \cdot F=e^{\int p d x}=e^{\int(-1) d x} \Rightarrow I \cdot F=e^{-x}\) Now for general solution- \(y \times I F=\int I F \times Q d x+c \Rightarrow y \cdot e^{-x}=\int x^{2} \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+\int 2 x \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+2\left[-\mathrm{xe}^{-x}-\left(e^{-x}\right)\right]+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+c\) \(\text { dividing by e } e^{-x} \text { then we get- }\) \(y=-x^{2}-2 x-2+e^{x}\)
AP EAMCET-2002
Differential Equation
87285
The general solution of \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\)
1 \(y-x^{2}=c \sec x\)
2 \(y \cos x=x^{2} \sec x+c\)
3 \(y \sec x=x^{2}+c \cos x\)
4 \(y=x^{2}+c \cos x\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\) Which is a linear differential equation \(P=\tan x \text { and } Q=2 x+x^{2} \tan x\) I.F \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int \tan \mathrm{x}} \Rightarrow=\mathrm{e}^{\log \sec x}=\sec \mathrm{x}\) For solution of equation - \(y \cdot \sec x=\int \sec x\left(2 x+x^{2} \tan x\right) d x+c\) \(y \cdot \sec x=\int\left(2 x \sec x+x^{2} \sec x \tan x\right) d x+c\) \(y \cdot \sec x=\int \frac{d}{d x}\left(x^{2} \sec x\right) d x+c\) \(y \sec x=x^{2} \sec x+c\) \(y=x^{2}+c \cos x\)
AP EAMCET-22.04.2018
Differential Equation
87271
The solution of the differential equation \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0\) is
1 \(\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0\)
2 \(\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0\)
3 \(\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0\)
4 \(\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
Explanation:
(D) : We have differential equation- \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right)\) Let \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) Then equation, \(v+x \frac{d v}{d x}=-\left(\frac{x^{2}+3 v^{2} x^{2}}{3 x^{2}+v^{2} x^{2}}\right)\) \(v+x \frac{d v}{d x}=-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=-v-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=\frac{-3 v-v^{3}-1-3 v^{2}}{3+v^{2}}\) \(x \frac{d v}{d x}=-\frac{(v+1)^{3}}{3+v^{2}}\) Separating the variable we get- \(\frac{3+v^{2}}{(v+1)^{3}} d v=-\frac{d x}{x}\) On integrating both side we get- \(\int \frac{3+v^{2}}{(v+1)^{3}} d v=-\int \frac{d x}{x}\) \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{A}{(v+1)}+\frac{B}{(v+1)^{2}}+\frac{C}{(v+1)^{3}}\) \(3+v^{2}=A\left(v^{2}+1+2 v\right)+B(v+1)+C\) \(A=1\) \(2 A+B=0, B=-2\) \(A+B+C=3\) \(1+(-2)+C=3\) \(\quad C=4\) Then, \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{1}{v+1}-\frac{2}{(v+1)^{2}}+\frac{4}{(v+1)^{3}}\) The equation becomes- \(-\int \frac{d x}{x}=\frac{1}{v+1} d v-\int \frac{2}{(v+1)^{2}} d v+\int \frac{4}{(v+1)^{3}} d v\) \(-\log x+\log c=\log (v+1)+\frac{2}{v+1}-\frac{4}{2(v+1)^{2}}\) Now substituting, \(v=y / x\) \(\log \mathrm{c} / \mathrm{x}=\log \left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)+\frac{2}{\frac{\mathrm{y}}{\mathrm{x}}+1}-\frac{2}{\left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)^{2}}\) \(\log \mathrm{c} / \mathrm{x}-\log \left(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{x}}\right)=\frac{2 \mathrm{x}^{2}+2 \mathrm{xy}-2 \mathrm{x}^{2}}{(\mathrm{x}+\mathrm{y})^{2}}\) \(\log \left(\frac{c}{x+y}\right)=\frac{2 x y}{(x+y)^{2}}\) \(-\log \left|\frac{x+y}{c}\right|=\frac{2 x y}{(x+y)^{2}}\) \(\log |x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
JEE Main-30.01.2023
Differential Equation
87272
The general solution of the differential equation \(\left(1+y^{2}\right) \mathbf{d x}+\left(1+\mathbf{x}^{2}\right) \mathbf{d y}=\mathbf{0}\) is
1 \(x-y=c(1-x y)\)
2 \(x-y=c(1-x y)\)
3 \(x+y=c(1-x y)\)
4 \(x+y=c(1+x y)\)
Explanation:
(C) : Differential equation \(\left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0\) On dividing of \(\left(1+x^{2}\right)\left(1+y^{2}\right)\) in above equation \(\frac{d x}{1+x^{2}}+\frac{d y}{1+y^{2}}=0\) On integrating both side, we get \(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}\) \(\frac{x+y}{1-x y}=c\) \(x+y=c(1-x y)\)
87270
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}=\frac{y}{x}\) \(\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3\). Then \(\frac{y^{2}(x)}{9}\) is equal to:
87284
\(y+x^{2}=\frac{d y}{d x}\) has the solution
1 \(y+x^{2}+2 x+2=c e^{x}\)
2 \(y+x+2 x^{2}+2=c e^{x}\)
3 \(y^{2}+x+x^{2}+2=c e^{2 x}\)
4 \(y+x+x^{2}+2=c e^{2 x}\)
Explanation:
(A) : We have- \(y+x^{2}=d y / d x\) It can be written as \(\mathrm{dy} / \mathrm{dx}-\mathrm{y}=\mathrm{x}^{2}\) Which is a linear differential equation. Where, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{x}^{2}\) \(I \cdot F=e^{\int p d x}=e^{\int(-1) d x} \Rightarrow I \cdot F=e^{-x}\) Now for general solution- \(y \times I F=\int I F \times Q d x+c \Rightarrow y \cdot e^{-x}=\int x^{2} \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+\int 2 x \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+2\left[-\mathrm{xe}^{-x}-\left(e^{-x}\right)\right]+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+c\) \(\text { dividing by e } e^{-x} \text { then we get- }\) \(y=-x^{2}-2 x-2+e^{x}\)
AP EAMCET-2002
Differential Equation
87285
The general solution of \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\)
1 \(y-x^{2}=c \sec x\)
2 \(y \cos x=x^{2} \sec x+c\)
3 \(y \sec x=x^{2}+c \cos x\)
4 \(y=x^{2}+c \cos x\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\) Which is a linear differential equation \(P=\tan x \text { and } Q=2 x+x^{2} \tan x\) I.F \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int \tan \mathrm{x}} \Rightarrow=\mathrm{e}^{\log \sec x}=\sec \mathrm{x}\) For solution of equation - \(y \cdot \sec x=\int \sec x\left(2 x+x^{2} \tan x\right) d x+c\) \(y \cdot \sec x=\int\left(2 x \sec x+x^{2} \sec x \tan x\right) d x+c\) \(y \cdot \sec x=\int \frac{d}{d x}\left(x^{2} \sec x\right) d x+c\) \(y \sec x=x^{2} \sec x+c\) \(y=x^{2}+c \cos x\)
AP EAMCET-22.04.2018
Differential Equation
87271
The solution of the differential equation \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0\) is
1 \(\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0\)
2 \(\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0\)
3 \(\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0\)
4 \(\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
Explanation:
(D) : We have differential equation- \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right)\) Let \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) Then equation, \(v+x \frac{d v}{d x}=-\left(\frac{x^{2}+3 v^{2} x^{2}}{3 x^{2}+v^{2} x^{2}}\right)\) \(v+x \frac{d v}{d x}=-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=-v-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=\frac{-3 v-v^{3}-1-3 v^{2}}{3+v^{2}}\) \(x \frac{d v}{d x}=-\frac{(v+1)^{3}}{3+v^{2}}\) Separating the variable we get- \(\frac{3+v^{2}}{(v+1)^{3}} d v=-\frac{d x}{x}\) On integrating both side we get- \(\int \frac{3+v^{2}}{(v+1)^{3}} d v=-\int \frac{d x}{x}\) \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{A}{(v+1)}+\frac{B}{(v+1)^{2}}+\frac{C}{(v+1)^{3}}\) \(3+v^{2}=A\left(v^{2}+1+2 v\right)+B(v+1)+C\) \(A=1\) \(2 A+B=0, B=-2\) \(A+B+C=3\) \(1+(-2)+C=3\) \(\quad C=4\) Then, \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{1}{v+1}-\frac{2}{(v+1)^{2}}+\frac{4}{(v+1)^{3}}\) The equation becomes- \(-\int \frac{d x}{x}=\frac{1}{v+1} d v-\int \frac{2}{(v+1)^{2}} d v+\int \frac{4}{(v+1)^{3}} d v\) \(-\log x+\log c=\log (v+1)+\frac{2}{v+1}-\frac{4}{2(v+1)^{2}}\) Now substituting, \(v=y / x\) \(\log \mathrm{c} / \mathrm{x}=\log \left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)+\frac{2}{\frac{\mathrm{y}}{\mathrm{x}}+1}-\frac{2}{\left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)^{2}}\) \(\log \mathrm{c} / \mathrm{x}-\log \left(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{x}}\right)=\frac{2 \mathrm{x}^{2}+2 \mathrm{xy}-2 \mathrm{x}^{2}}{(\mathrm{x}+\mathrm{y})^{2}}\) \(\log \left(\frac{c}{x+y}\right)=\frac{2 x y}{(x+y)^{2}}\) \(-\log \left|\frac{x+y}{c}\right|=\frac{2 x y}{(x+y)^{2}}\) \(\log |x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
JEE Main-30.01.2023
Differential Equation
87272
The general solution of the differential equation \(\left(1+y^{2}\right) \mathbf{d x}+\left(1+\mathbf{x}^{2}\right) \mathbf{d y}=\mathbf{0}\) is
1 \(x-y=c(1-x y)\)
2 \(x-y=c(1-x y)\)
3 \(x+y=c(1-x y)\)
4 \(x+y=c(1+x y)\)
Explanation:
(C) : Differential equation \(\left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0\) On dividing of \(\left(1+x^{2}\right)\left(1+y^{2}\right)\) in above equation \(\frac{d x}{1+x^{2}}+\frac{d y}{1+y^{2}}=0\) On integrating both side, we get \(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}\) \(\frac{x+y}{1-x y}=c\) \(x+y=c(1-x y)\)
87270
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}=\frac{y}{x}\) \(\left(1+x y^{2}\left(1+\log _{e} x\right)\right), x>0, y(1)=3\). Then \(\frac{y^{2}(x)}{9}\) is equal to:
87284
\(y+x^{2}=\frac{d y}{d x}\) has the solution
1 \(y+x^{2}+2 x+2=c e^{x}\)
2 \(y+x+2 x^{2}+2=c e^{x}\)
3 \(y^{2}+x+x^{2}+2=c e^{2 x}\)
4 \(y+x+x^{2}+2=c e^{2 x}\)
Explanation:
(A) : We have- \(y+x^{2}=d y / d x\) It can be written as \(\mathrm{dy} / \mathrm{dx}-\mathrm{y}=\mathrm{x}^{2}\) Which is a linear differential equation. Where, \(\mathrm{P}=-1, \mathrm{Q}=\mathrm{x}^{2}\) \(I \cdot F=e^{\int p d x}=e^{\int(-1) d x} \Rightarrow I \cdot F=e^{-x}\) Now for general solution- \(y \times I F=\int I F \times Q d x+c \Rightarrow y \cdot e^{-x}=\int x^{2} \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+\int 2 x \cdot e^{-x} d x+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}+2\left[-\mathrm{xe}^{-x}-\left(e^{-x}\right)\right]+c\) \(y \cdot e^{-x}=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+c\) \(\text { dividing by e } e^{-x} \text { then we get- }\) \(y=-x^{2}-2 x-2+e^{x}\)
AP EAMCET-2002
Differential Equation
87285
The general solution of \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\)
1 \(y-x^{2}=c \sec x\)
2 \(y \cos x=x^{2} \sec x+c\)
3 \(y \sec x=x^{2}+c \cos x\)
4 \(y=x^{2}+c \cos x\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x\) Which is a linear differential equation \(P=\tan x \text { and } Q=2 x+x^{2} \tan x\) I.F \(\quad=\mathrm{e}^{\int \mathrm{pdx}} \Rightarrow=\mathrm{e}^{\int \tan \mathrm{x}} \Rightarrow=\mathrm{e}^{\log \sec x}=\sec \mathrm{x}\) For solution of equation - \(y \cdot \sec x=\int \sec x\left(2 x+x^{2} \tan x\right) d x+c\) \(y \cdot \sec x=\int\left(2 x \sec x+x^{2} \sec x \tan x\right) d x+c\) \(y \cdot \sec x=\int \frac{d}{d x}\left(x^{2} \sec x\right) d x+c\) \(y \sec x=x^{2} \sec x+c\) \(y=x^{2}+c \cos x\)
AP EAMCET-22.04.2018
Differential Equation
87271
The solution of the differential equation \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right), y(1)=0\) is
1 \(\log _{e}|x+y|-\frac{x y}{(x+y)^{2}}=0\)
2 \(\log _{e}|x+y|-\frac{2 x y}{(x+y)^{2}}=0\)
3 \(\log _{e}|x+y|+\frac{x y}{(x+y)^{2}}=0\)
4 \(\log _{e}|x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
Explanation:
(D) : We have differential equation- \(\frac{d y}{d x}=-\left(\frac{x^{2}+3 y^{2}}{3 x^{2}+y^{2}}\right)\) Let \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) Then equation, \(v+x \frac{d v}{d x}=-\left(\frac{x^{2}+3 v^{2} x^{2}}{3 x^{2}+v^{2} x^{2}}\right)\) \(v+x \frac{d v}{d x}=-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=-v-\left(\frac{1+3 v^{2}}{3+v^{2}}\right)\) \(x \frac{d v}{d x}=\frac{-3 v-v^{3}-1-3 v^{2}}{3+v^{2}}\) \(x \frac{d v}{d x}=-\frac{(v+1)^{3}}{3+v^{2}}\) Separating the variable we get- \(\frac{3+v^{2}}{(v+1)^{3}} d v=-\frac{d x}{x}\) On integrating both side we get- \(\int \frac{3+v^{2}}{(v+1)^{3}} d v=-\int \frac{d x}{x}\) \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{A}{(v+1)}+\frac{B}{(v+1)^{2}}+\frac{C}{(v+1)^{3}}\) \(3+v^{2}=A\left(v^{2}+1+2 v\right)+B(v+1)+C\) \(A=1\) \(2 A+B=0, B=-2\) \(A+B+C=3\) \(1+(-2)+C=3\) \(\quad C=4\) Then, \(\frac{3+v^{2}}{(v+1)^{3}}=\frac{1}{v+1}-\frac{2}{(v+1)^{2}}+\frac{4}{(v+1)^{3}}\) The equation becomes- \(-\int \frac{d x}{x}=\frac{1}{v+1} d v-\int \frac{2}{(v+1)^{2}} d v+\int \frac{4}{(v+1)^{3}} d v\) \(-\log x+\log c=\log (v+1)+\frac{2}{v+1}-\frac{4}{2(v+1)^{2}}\) Now substituting, \(v=y / x\) \(\log \mathrm{c} / \mathrm{x}=\log \left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)+\frac{2}{\frac{\mathrm{y}}{\mathrm{x}}+1}-\frac{2}{\left(\frac{\mathrm{y}}{\mathrm{x}}+1\right)^{2}}\) \(\log \mathrm{c} / \mathrm{x}-\log \left(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{x}}\right)=\frac{2 \mathrm{x}^{2}+2 \mathrm{xy}-2 \mathrm{x}^{2}}{(\mathrm{x}+\mathrm{y})^{2}}\) \(\log \left(\frac{c}{x+y}\right)=\frac{2 x y}{(x+y)^{2}}\) \(-\log \left|\frac{x+y}{c}\right|=\frac{2 x y}{(x+y)^{2}}\) \(\log |x+y|+\frac{2 x y}{(x+y)^{2}}=0\)
JEE Main-30.01.2023
Differential Equation
87272
The general solution of the differential equation \(\left(1+y^{2}\right) \mathbf{d x}+\left(1+\mathbf{x}^{2}\right) \mathbf{d y}=\mathbf{0}\) is
1 \(x-y=c(1-x y)\)
2 \(x-y=c(1-x y)\)
3 \(x+y=c(1-x y)\)
4 \(x+y=c(1+x y)\)
Explanation:
(C) : Differential equation \(\left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0\) On dividing of \(\left(1+x^{2}\right)\left(1+y^{2}\right)\) in above equation \(\frac{d x}{1+x^{2}}+\frac{d y}{1+y^{2}}=0\) On integrating both side, we get \(\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\mathrm{c}\) \(\frac{x+y}{1-x y}=c\) \(x+y=c(1-x y)\)
