(A) : We have differential equation - \(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \Rightarrow \frac{d y}{d x}=e^{x} e^{-y}+x^{2} e^{-y}\) \(\frac{d y}{d x}=e^{-y}\left(e^{x}+x^{2}\right)\) Integrating on both sides, we get- \(\int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x\) \(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)
UPSEE-2006
Differential Equation
87209
Solution of the differential equation \(y d x+\left(x-y^{2}\right) d y=0\) is
1 \(x y+y^{3}=3 \mathrm{c}\)
2 \(x y-y^{3}=3 c\)
3 \(3 x y-y^{3}=3 c\)
4 None of these
Explanation:
(C) : Given that that differential equation - \(y d x+\left(x-y^{2}\right) d y=0 \Rightarrow y \frac{d x}{d y}+x-y^{2}=0\) \(\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{y}\) This is of the form \(\frac{d x}{d y}+P x=Q\) Where, \(\mathrm{P}=\frac{1}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}\) I.F. \(=\mathrm{e}^{\int \text { Pdy }}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log y}=\mathrm{y}\) \(\therefore\) The solution is given by \(x y=\int y \cdot y d y+c \Rightarrow x y=\frac{y^{3}}{3}+c\) \(3 x y-y^{3}=3 \mathrm{c}\)
COMEDK-2020
Differential Equation
87210
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) is
1 \(y\left(1+x^{2}\right)=C+\tan ^{-1} x\)
2 \(\frac{y}{1+x^{2}}=C+\tan ^{-1} x\)
3 \(y \log \left(1+x^{2}\right)=C+\tan ^{-1} x\)
4 \(y\left(1+x^{2}\right)=C+\sin ^{-1} x\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) Here, \(P=\frac{2 x}{1+x^{2}}, \quad Q=\frac{1}{\left(1+x^{2}\right)^{2}}\) Now, I.F. \(=\mathrm{e}^{\int \text { Pdx }}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^{2}\right)}\) I.F. \(=1+x^{2}\) Solution of differential equation is \(\mathrm{y} \cdot \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF} \mathrm{dx}+\mathrm{C}\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right) d x+C\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)} d x+C\) \(y\left(1+x^{2}\right)=\tan ^{-1} x+C\)
VITEEE-2012
Differential Equation
87211
The integrating factor of \(\frac{x d y}{d x}-y=x^{4}-3 x\) is
1 \(x\)
2 \(\log \mathrm{x}\)
3 \(\frac{1}{x}\)
4 \(-\mathrm{X}\)
Explanation:
(C) : Given that the differential equation - \(x \frac{d y}{d x}-y=x^{4}-3 x\) Dividing by \(\mathrm{x}\) on both side, we get - It is form of \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{3}-3\) \(\frac{d y}{d x}+P y=Q\) Here, \(\quad \mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) Hence, \(\mathrm{I} \cdot \mathrm{F}=\mathrm{e}^{\int \mathrm{Pdx}} \Rightarrow=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log x}\) \(\mathrm{I} \cdot \mathrm{F}=\frac{1}{\mathrm{x}}\)
Differential Equation
87212
The solution of the differential equation \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) is
(C) : Given that, \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) Rearranging the equation, we have \(d x-y d y+\sqrt{\left(x^{2}+y^{2}\right)}(x d x+y d y)=0\) \(d x-y d y+\frac{1}{2} \sqrt{\left(x^{2}+y^{2}\right)}(2 x d x+2 y d y)=0\) Let \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{t}\) \(2 x d x+2 y d y=d t\) Integrating both side, we get - \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt}=\mathrm{C} \Rightarrow \mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \times \frac{\mathrm{t}^{3 / 2}}{3 / 2}=\mathrm{C}\) Putting the value of \(t=x^{2}+y^{2}\) \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{3}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{3 / 2}=\mathrm{C}\)
(A) : We have differential equation - \(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \Rightarrow \frac{d y}{d x}=e^{x} e^{-y}+x^{2} e^{-y}\) \(\frac{d y}{d x}=e^{-y}\left(e^{x}+x^{2}\right)\) Integrating on both sides, we get- \(\int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x\) \(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)
UPSEE-2006
Differential Equation
87209
Solution of the differential equation \(y d x+\left(x-y^{2}\right) d y=0\) is
1 \(x y+y^{3}=3 \mathrm{c}\)
2 \(x y-y^{3}=3 c\)
3 \(3 x y-y^{3}=3 c\)
4 None of these
Explanation:
(C) : Given that that differential equation - \(y d x+\left(x-y^{2}\right) d y=0 \Rightarrow y \frac{d x}{d y}+x-y^{2}=0\) \(\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{y}\) This is of the form \(\frac{d x}{d y}+P x=Q\) Where, \(\mathrm{P}=\frac{1}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}\) I.F. \(=\mathrm{e}^{\int \text { Pdy }}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log y}=\mathrm{y}\) \(\therefore\) The solution is given by \(x y=\int y \cdot y d y+c \Rightarrow x y=\frac{y^{3}}{3}+c\) \(3 x y-y^{3}=3 \mathrm{c}\)
COMEDK-2020
Differential Equation
87210
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) is
1 \(y\left(1+x^{2}\right)=C+\tan ^{-1} x\)
2 \(\frac{y}{1+x^{2}}=C+\tan ^{-1} x\)
3 \(y \log \left(1+x^{2}\right)=C+\tan ^{-1} x\)
4 \(y\left(1+x^{2}\right)=C+\sin ^{-1} x\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) Here, \(P=\frac{2 x}{1+x^{2}}, \quad Q=\frac{1}{\left(1+x^{2}\right)^{2}}\) Now, I.F. \(=\mathrm{e}^{\int \text { Pdx }}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^{2}\right)}\) I.F. \(=1+x^{2}\) Solution of differential equation is \(\mathrm{y} \cdot \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF} \mathrm{dx}+\mathrm{C}\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right) d x+C\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)} d x+C\) \(y\left(1+x^{2}\right)=\tan ^{-1} x+C\)
VITEEE-2012
Differential Equation
87211
The integrating factor of \(\frac{x d y}{d x}-y=x^{4}-3 x\) is
1 \(x\)
2 \(\log \mathrm{x}\)
3 \(\frac{1}{x}\)
4 \(-\mathrm{X}\)
Explanation:
(C) : Given that the differential equation - \(x \frac{d y}{d x}-y=x^{4}-3 x\) Dividing by \(\mathrm{x}\) on both side, we get - It is form of \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{3}-3\) \(\frac{d y}{d x}+P y=Q\) Here, \(\quad \mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) Hence, \(\mathrm{I} \cdot \mathrm{F}=\mathrm{e}^{\int \mathrm{Pdx}} \Rightarrow=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log x}\) \(\mathrm{I} \cdot \mathrm{F}=\frac{1}{\mathrm{x}}\)
Differential Equation
87212
The solution of the differential equation \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) is
(C) : Given that, \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) Rearranging the equation, we have \(d x-y d y+\sqrt{\left(x^{2}+y^{2}\right)}(x d x+y d y)=0\) \(d x-y d y+\frac{1}{2} \sqrt{\left(x^{2}+y^{2}\right)}(2 x d x+2 y d y)=0\) Let \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{t}\) \(2 x d x+2 y d y=d t\) Integrating both side, we get - \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt}=\mathrm{C} \Rightarrow \mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \times \frac{\mathrm{t}^{3 / 2}}{3 / 2}=\mathrm{C}\) Putting the value of \(t=x^{2}+y^{2}\) \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{3}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{3 / 2}=\mathrm{C}\)
(A) : We have differential equation - \(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \Rightarrow \frac{d y}{d x}=e^{x} e^{-y}+x^{2} e^{-y}\) \(\frac{d y}{d x}=e^{-y}\left(e^{x}+x^{2}\right)\) Integrating on both sides, we get- \(\int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x\) \(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)
UPSEE-2006
Differential Equation
87209
Solution of the differential equation \(y d x+\left(x-y^{2}\right) d y=0\) is
1 \(x y+y^{3}=3 \mathrm{c}\)
2 \(x y-y^{3}=3 c\)
3 \(3 x y-y^{3}=3 c\)
4 None of these
Explanation:
(C) : Given that that differential equation - \(y d x+\left(x-y^{2}\right) d y=0 \Rightarrow y \frac{d x}{d y}+x-y^{2}=0\) \(\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{y}\) This is of the form \(\frac{d x}{d y}+P x=Q\) Where, \(\mathrm{P}=\frac{1}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}\) I.F. \(=\mathrm{e}^{\int \text { Pdy }}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log y}=\mathrm{y}\) \(\therefore\) The solution is given by \(x y=\int y \cdot y d y+c \Rightarrow x y=\frac{y^{3}}{3}+c\) \(3 x y-y^{3}=3 \mathrm{c}\)
COMEDK-2020
Differential Equation
87210
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) is
1 \(y\left(1+x^{2}\right)=C+\tan ^{-1} x\)
2 \(\frac{y}{1+x^{2}}=C+\tan ^{-1} x\)
3 \(y \log \left(1+x^{2}\right)=C+\tan ^{-1} x\)
4 \(y\left(1+x^{2}\right)=C+\sin ^{-1} x\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) Here, \(P=\frac{2 x}{1+x^{2}}, \quad Q=\frac{1}{\left(1+x^{2}\right)^{2}}\) Now, I.F. \(=\mathrm{e}^{\int \text { Pdx }}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^{2}\right)}\) I.F. \(=1+x^{2}\) Solution of differential equation is \(\mathrm{y} \cdot \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF} \mathrm{dx}+\mathrm{C}\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right) d x+C\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)} d x+C\) \(y\left(1+x^{2}\right)=\tan ^{-1} x+C\)
VITEEE-2012
Differential Equation
87211
The integrating factor of \(\frac{x d y}{d x}-y=x^{4}-3 x\) is
1 \(x\)
2 \(\log \mathrm{x}\)
3 \(\frac{1}{x}\)
4 \(-\mathrm{X}\)
Explanation:
(C) : Given that the differential equation - \(x \frac{d y}{d x}-y=x^{4}-3 x\) Dividing by \(\mathrm{x}\) on both side, we get - It is form of \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{3}-3\) \(\frac{d y}{d x}+P y=Q\) Here, \(\quad \mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) Hence, \(\mathrm{I} \cdot \mathrm{F}=\mathrm{e}^{\int \mathrm{Pdx}} \Rightarrow=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log x}\) \(\mathrm{I} \cdot \mathrm{F}=\frac{1}{\mathrm{x}}\)
Differential Equation
87212
The solution of the differential equation \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) is
(C) : Given that, \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) Rearranging the equation, we have \(d x-y d y+\sqrt{\left(x^{2}+y^{2}\right)}(x d x+y d y)=0\) \(d x-y d y+\frac{1}{2} \sqrt{\left(x^{2}+y^{2}\right)}(2 x d x+2 y d y)=0\) Let \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{t}\) \(2 x d x+2 y d y=d t\) Integrating both side, we get - \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt}=\mathrm{C} \Rightarrow \mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \times \frac{\mathrm{t}^{3 / 2}}{3 / 2}=\mathrm{C}\) Putting the value of \(t=x^{2}+y^{2}\) \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{3}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{3 / 2}=\mathrm{C}\)
(A) : We have differential equation - \(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \Rightarrow \frac{d y}{d x}=e^{x} e^{-y}+x^{2} e^{-y}\) \(\frac{d y}{d x}=e^{-y}\left(e^{x}+x^{2}\right)\) Integrating on both sides, we get- \(\int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x\) \(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)
UPSEE-2006
Differential Equation
87209
Solution of the differential equation \(y d x+\left(x-y^{2}\right) d y=0\) is
1 \(x y+y^{3}=3 \mathrm{c}\)
2 \(x y-y^{3}=3 c\)
3 \(3 x y-y^{3}=3 c\)
4 None of these
Explanation:
(C) : Given that that differential equation - \(y d x+\left(x-y^{2}\right) d y=0 \Rightarrow y \frac{d x}{d y}+x-y^{2}=0\) \(\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{y}\) This is of the form \(\frac{d x}{d y}+P x=Q\) Where, \(\mathrm{P}=\frac{1}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}\) I.F. \(=\mathrm{e}^{\int \text { Pdy }}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log y}=\mathrm{y}\) \(\therefore\) The solution is given by \(x y=\int y \cdot y d y+c \Rightarrow x y=\frac{y^{3}}{3}+c\) \(3 x y-y^{3}=3 \mathrm{c}\)
COMEDK-2020
Differential Equation
87210
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) is
1 \(y\left(1+x^{2}\right)=C+\tan ^{-1} x\)
2 \(\frac{y}{1+x^{2}}=C+\tan ^{-1} x\)
3 \(y \log \left(1+x^{2}\right)=C+\tan ^{-1} x\)
4 \(y\left(1+x^{2}\right)=C+\sin ^{-1} x\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) Here, \(P=\frac{2 x}{1+x^{2}}, \quad Q=\frac{1}{\left(1+x^{2}\right)^{2}}\) Now, I.F. \(=\mathrm{e}^{\int \text { Pdx }}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^{2}\right)}\) I.F. \(=1+x^{2}\) Solution of differential equation is \(\mathrm{y} \cdot \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF} \mathrm{dx}+\mathrm{C}\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right) d x+C\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)} d x+C\) \(y\left(1+x^{2}\right)=\tan ^{-1} x+C\)
VITEEE-2012
Differential Equation
87211
The integrating factor of \(\frac{x d y}{d x}-y=x^{4}-3 x\) is
1 \(x\)
2 \(\log \mathrm{x}\)
3 \(\frac{1}{x}\)
4 \(-\mathrm{X}\)
Explanation:
(C) : Given that the differential equation - \(x \frac{d y}{d x}-y=x^{4}-3 x\) Dividing by \(\mathrm{x}\) on both side, we get - It is form of \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{3}-3\) \(\frac{d y}{d x}+P y=Q\) Here, \(\quad \mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) Hence, \(\mathrm{I} \cdot \mathrm{F}=\mathrm{e}^{\int \mathrm{Pdx}} \Rightarrow=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log x}\) \(\mathrm{I} \cdot \mathrm{F}=\frac{1}{\mathrm{x}}\)
Differential Equation
87212
The solution of the differential equation \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) is
(C) : Given that, \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) Rearranging the equation, we have \(d x-y d y+\sqrt{\left(x^{2}+y^{2}\right)}(x d x+y d y)=0\) \(d x-y d y+\frac{1}{2} \sqrt{\left(x^{2}+y^{2}\right)}(2 x d x+2 y d y)=0\) Let \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{t}\) \(2 x d x+2 y d y=d t\) Integrating both side, we get - \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt}=\mathrm{C} \Rightarrow \mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \times \frac{\mathrm{t}^{3 / 2}}{3 / 2}=\mathrm{C}\) Putting the value of \(t=x^{2}+y^{2}\) \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{3}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{3 / 2}=\mathrm{C}\)
(A) : We have differential equation - \(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y} \Rightarrow \frac{d y}{d x}=e^{x} e^{-y}+x^{2} e^{-y}\) \(\frac{d y}{d x}=e^{-y}\left(e^{x}+x^{2}\right)\) Integrating on both sides, we get- \(\int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x\) \(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)
UPSEE-2006
Differential Equation
87209
Solution of the differential equation \(y d x+\left(x-y^{2}\right) d y=0\) is
1 \(x y+y^{3}=3 \mathrm{c}\)
2 \(x y-y^{3}=3 c\)
3 \(3 x y-y^{3}=3 c\)
4 None of these
Explanation:
(C) : Given that that differential equation - \(y d x+\left(x-y^{2}\right) d y=0 \Rightarrow y \frac{d x}{d y}+x-y^{2}=0\) \(\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{y}\) This is of the form \(\frac{d x}{d y}+P x=Q\) Where, \(\mathrm{P}=\frac{1}{\mathrm{y}}, \mathrm{Q}=\mathrm{y}\) I.F. \(=\mathrm{e}^{\int \text { Pdy }}=\mathrm{e}^{\int \frac{\mathrm{dy}}{\mathrm{y}}}=\mathrm{e}^{\log y}=\mathrm{y}\) \(\therefore\) The solution is given by \(x y=\int y \cdot y d y+c \Rightarrow x y=\frac{y^{3}}{3}+c\) \(3 x y-y^{3}=3 \mathrm{c}\)
COMEDK-2020
Differential Equation
87210
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) is
1 \(y\left(1+x^{2}\right)=C+\tan ^{-1} x\)
2 \(\frac{y}{1+x^{2}}=C+\tan ^{-1} x\)
3 \(y \log \left(1+x^{2}\right)=C+\tan ^{-1} x\)
4 \(y\left(1+x^{2}\right)=C+\sin ^{-1} x\)
Explanation:
(A) : Given differential equation, \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) Here, \(P=\frac{2 x}{1+x^{2}}, \quad Q=\frac{1}{\left(1+x^{2}\right)^{2}}\) Now, I.F. \(=\mathrm{e}^{\int \text { Pdx }}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^{2}\right)}\) I.F. \(=1+x^{2}\) Solution of differential equation is \(\mathrm{y} \cdot \mathrm{IF}=\int \mathrm{Q} \cdot \mathrm{IF} \mathrm{dx}+\mathrm{C}\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}} \cdot\left(1+x^{2}\right) d x+C\) \(y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)} d x+C\) \(y\left(1+x^{2}\right)=\tan ^{-1} x+C\)
VITEEE-2012
Differential Equation
87211
The integrating factor of \(\frac{x d y}{d x}-y=x^{4}-3 x\) is
1 \(x\)
2 \(\log \mathrm{x}\)
3 \(\frac{1}{x}\)
4 \(-\mathrm{X}\)
Explanation:
(C) : Given that the differential equation - \(x \frac{d y}{d x}-y=x^{4}-3 x\) Dividing by \(\mathrm{x}\) on both side, we get - It is form of \(\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{3}-3\) \(\frac{d y}{d x}+P y=Q\) Here, \(\quad \mathrm{P}=\frac{1}{\mathrm{x}}\) and \(\mathrm{Q}=\mathrm{x}^{3}-3\) Hence, \(\mathrm{I} \cdot \mathrm{F}=\mathrm{e}^{\int \mathrm{Pdx}} \Rightarrow=\mathrm{e}^{-\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-\log x}\) \(\mathrm{I} \cdot \mathrm{F}=\frac{1}{\mathrm{x}}\)
Differential Equation
87212
The solution of the differential equation \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) is
(C) : Given that, \(\left\{1+x \sqrt{\left(x^{2}+y^{2}\right)}\right\} d x+\left\{\sqrt{\left(x^{2}+y^{2}\right)}-1\right\} y d y=0\) Rearranging the equation, we have \(d x-y d y+\sqrt{\left(x^{2}+y^{2}\right)}(x d x+y d y)=0\) \(d x-y d y+\frac{1}{2} \sqrt{\left(x^{2}+y^{2}\right)}(2 x d x+2 y d y)=0\) Let \(\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{t}\) \(2 x d x+2 y d y=d t\) Integrating both side, we get - \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \int \sqrt{\mathrm{t}} \mathrm{dt}=\mathrm{C} \Rightarrow \mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{2} \times \frac{\mathrm{t}^{3 / 2}}{3 / 2}=\mathrm{C}\) Putting the value of \(t=x^{2}+y^{2}\) \(\mathrm{x}-\frac{\mathrm{y}^{2}}{2}+\frac{1}{3}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{3 / 2}=\mathrm{C}\)