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Differential Equation

87323 The general solution of the differential equation \(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\) is \(\quad\) (c is an arbitrary constant)

1 \(x-y e^{\frac{x}{y}}=c\)

2 \(y-x e^{\frac{x}{y}}=c\)

3 \(x y^{\frac{x}{y}} c\)

4 \(y+x e^{\frac{x}{y}}=c\)

Explanation:

(C) : Given,
\(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\)
\(\Rightarrow\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}=-\mathrm{e}^{\mathrm{x} / \mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}\)
\(\Rightarrow \frac{d x}{d y}=-\frac{e^{x / y}}{\left(1+e^{x / y}\right)}\left(1-\frac{x}{y}\right)\)
This is homogeneous differential equation.
So, put \(\mathrm{x}=\mathrm{vy}\)
\(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
Therefore, \(v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}\)
[From equation (i) and (ii)]
\(\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}-\mathrm{v}\)
\(=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}\left(1+\mathrm{e}^{\mathrm{v}}\right)}{1+\mathrm{e}^{\mathrm{v}}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}-\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}\)
\(y \frac{d v}{d y}=\frac{-\left(e^{v}+v\right)}{1+e^{v}}\)
\(\Rightarrow \frac{1+e^{v}}{v+e^{v}} d v=-\frac{d y}{y} \quad\) [Variable separation]
\(\Rightarrow \int \frac{1+e^{v}}{v+e^{v}} d v=-\int \frac{d y}{y}\)
Let \(\mathrm{v}+\mathrm{e}^{\mathrm{v}}=\mathrm{t} \Rightarrow\left(1+\mathrm{e}^{\mathrm{v}}\right) \mathrm{dy}=\mathrm{dt}\)
\(\therefore \int \frac{1+\mathrm{e}^{\mathrm{v}}}{\mathrm{v}+\mathrm{e}^{\mathrm{v}}} \mathrm{dv}=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log \mathrm{t}=\log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)\)
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=-\log \mathrm{y}+\log \mathrm{C} \quad\) [From equation (iii) and (iv)]
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)+\log \mathrm{y}=\log \mathrm{C}\)
\(\Rightarrow \mathrm{y}\left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=\mathrm{C}\)
\(\Rightarrow y\left(\frac{x}{y}+e^{x / y}\right)=C \Rightarrow y\left(\frac{x+y e^{x / y}}{y}\right)=C\)
\(\Rightarrow \mathrm{x}+\mathrm{ye}^{\mathrm{x} / \mathrm{y}}=\mathrm{C}\)

WB JEE-2019

Differential Equation

87324 General solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0\) is ( \(C\) is an arbitrary constant)

1 \(\frac{x}{a}=\tan \frac{y}{a}+c\)

2 \(\tan x y=c\)

3 \(\tan (x+y)=c\)

4 \(\tan \frac{y+c}{a}=\frac{x+y}{a}\)

Explanation:

(D) : Given,
Let,
\((x+y)^{2} \frac{d y}{d x}=a^{2}, a=0\)
et, \(\quad x+y=t\)
\(1+\frac{d y}{d x}=\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1\)
Therefore, \(\mathrm{t}^{2}\left(\frac{\mathrm{dt}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\)
\(\Rightarrow \quad \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{a}^{2}+\mathrm{t}^{2}\)
\(\frac{\mathrm{t}^{2}}{\mathrm{a}^{2}+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dx} \quad[\text { Variable separation] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\int \mathrm{dx} \quad \text { [int egration] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}+\mathrm{a}^{2}-\mathrm{a}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int\left(1-\frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}}\right) \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int \mathrm{dt}-\int \frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\mathrm{a}^{2}}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2}\left[\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right]=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow(\mathrm{x}+\mathrm{y})-\mathrm{a} \tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}\right)\)
\(=\mathrm{x}+\mathrm{C}^{\prime}[\) on putting the valueof \(\mathrm{t}]\)
\(\Rightarrow \frac{y-C^{\prime}}{a}=\tan ^{-1}\left(\frac{x+y}{a}\right)\)
\(\Rightarrow \frac{x+y}{a}=\tan \left(\frac{y-C^{\prime}}{a}\right) \Rightarrow \frac{x+y}{a}=\tan \left(\frac{y+C}{a}\right)\)
[where, \(\mathrm{C}=-\mathrm{C}^{\prime}\) is an arbitrary constant]
\(\tan \frac{\mathrm{y} C}{\mathrm{a}} \frac{\mathrm{x} y}{\mathrm{a}}\)

WB JEE-2019

Differential Equation

87325 The differential equation formed by eliminating \(a\) and \(b\) from the equation \(y=e^{x}(\operatorname{acos} x+b \sin x)\) is

1 \(2 \frac{d^{2} y}{d^{2}}+\frac{d y}{d x}-2 y=0\)

2 \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-2 y=0\)

3 \(2 \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}+2 y=0\)

4 \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\)

AP EAMCET-21.04.2019

Differential Equation

87327 The solution of the differential equation \(y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y\) satisfying \(y\left(\frac{\pi}{4}\right)=1\) is

1 \(\cos \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

2 \(\sin \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

3 \(\sin \frac{x}{y}=\log _{e} x-\frac{1}{\sqrt{2}}\)

4 \(\cos \frac{x}{y}=-\log _{e} x-\frac{1}{\sqrt{2}}\)

WB JEE-2013

Differential Equation

87323 The general solution of the differential equation \(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\) is \(\quad\) (c is an arbitrary constant)

1 \(x-y e^{\frac{x}{y}}=c\)

2 \(y-x e^{\frac{x}{y}}=c\)

3 \(x y^{\frac{x}{y}} c\)

4 \(y+x e^{\frac{x}{y}}=c\)

Explanation:

(C) : Given,
\(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\)
\(\Rightarrow\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}=-\mathrm{e}^{\mathrm{x} / \mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}\)
\(\Rightarrow \frac{d x}{d y}=-\frac{e^{x / y}}{\left(1+e^{x / y}\right)}\left(1-\frac{x}{y}\right)\)
This is homogeneous differential equation.
So, put \(\mathrm{x}=\mathrm{vy}\)
\(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
Therefore, \(v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}\)
[From equation (i) and (ii)]
\(\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}-\mathrm{v}\)
\(=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}\left(1+\mathrm{e}^{\mathrm{v}}\right)}{1+\mathrm{e}^{\mathrm{v}}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}-\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}\)
\(y \frac{d v}{d y}=\frac{-\left(e^{v}+v\right)}{1+e^{v}}\)
\(\Rightarrow \frac{1+e^{v}}{v+e^{v}} d v=-\frac{d y}{y} \quad\) [Variable separation]
\(\Rightarrow \int \frac{1+e^{v}}{v+e^{v}} d v=-\int \frac{d y}{y}\)
Let \(\mathrm{v}+\mathrm{e}^{\mathrm{v}}=\mathrm{t} \Rightarrow\left(1+\mathrm{e}^{\mathrm{v}}\right) \mathrm{dy}=\mathrm{dt}\)
\(\therefore \int \frac{1+\mathrm{e}^{\mathrm{v}}}{\mathrm{v}+\mathrm{e}^{\mathrm{v}}} \mathrm{dv}=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log \mathrm{t}=\log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)\)
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=-\log \mathrm{y}+\log \mathrm{C} \quad\) [From equation (iii) and (iv)]
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)+\log \mathrm{y}=\log \mathrm{C}\)
\(\Rightarrow \mathrm{y}\left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=\mathrm{C}\)
\(\Rightarrow y\left(\frac{x}{y}+e^{x / y}\right)=C \Rightarrow y\left(\frac{x+y e^{x / y}}{y}\right)=C\)
\(\Rightarrow \mathrm{x}+\mathrm{ye}^{\mathrm{x} / \mathrm{y}}=\mathrm{C}\)

WB JEE-2019

Differential Equation

87324 General solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0\) is ( \(C\) is an arbitrary constant)

1 \(\frac{x}{a}=\tan \frac{y}{a}+c\)

2 \(\tan x y=c\)

3 \(\tan (x+y)=c\)

4 \(\tan \frac{y+c}{a}=\frac{x+y}{a}\)

Explanation:

(D) : Given,
Let,
\((x+y)^{2} \frac{d y}{d x}=a^{2}, a=0\)
et, \(\quad x+y=t\)
\(1+\frac{d y}{d x}=\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1\)
Therefore, \(\mathrm{t}^{2}\left(\frac{\mathrm{dt}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\)
\(\Rightarrow \quad \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{a}^{2}+\mathrm{t}^{2}\)
\(\frac{\mathrm{t}^{2}}{\mathrm{a}^{2}+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dx} \quad[\text { Variable separation] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\int \mathrm{dx} \quad \text { [int egration] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}+\mathrm{a}^{2}-\mathrm{a}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int\left(1-\frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}}\right) \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int \mathrm{dt}-\int \frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\mathrm{a}^{2}}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2}\left[\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right]=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow(\mathrm{x}+\mathrm{y})-\mathrm{a} \tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}\right)\)
\(=\mathrm{x}+\mathrm{C}^{\prime}[\) on putting the valueof \(\mathrm{t}]\)
\(\Rightarrow \frac{y-C^{\prime}}{a}=\tan ^{-1}\left(\frac{x+y}{a}\right)\)
\(\Rightarrow \frac{x+y}{a}=\tan \left(\frac{y-C^{\prime}}{a}\right) \Rightarrow \frac{x+y}{a}=\tan \left(\frac{y+C}{a}\right)\)
[where, \(\mathrm{C}=-\mathrm{C}^{\prime}\) is an arbitrary constant]
\(\tan \frac{\mathrm{y} C}{\mathrm{a}} \frac{\mathrm{x} y}{\mathrm{a}}\)

WB JEE-2019

Differential Equation

87325 The differential equation formed by eliminating \(a\) and \(b\) from the equation \(y=e^{x}(\operatorname{acos} x+b \sin x)\) is

1 \(2 \frac{d^{2} y}{d^{2}}+\frac{d y}{d x}-2 y=0\)

2 \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-2 y=0\)

3 \(2 \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}+2 y=0\)

4 \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\)

AP EAMCET-21.04.2019

Differential Equation

87327 The solution of the differential equation \(y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y\) satisfying \(y\left(\frac{\pi}{4}\right)=1\) is

1 \(\cos \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

2 \(\sin \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

3 \(\sin \frac{x}{y}=\log _{e} x-\frac{1}{\sqrt{2}}\)

4 \(\cos \frac{x}{y}=-\log _{e} x-\frac{1}{\sqrt{2}}\)

WB JEE-2013

Differential Equation

87323 The general solution of the differential equation \(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\) is \(\quad\) (c is an arbitrary constant)

1 \(x-y e^{\frac{x}{y}}=c\)

2 \(y-x e^{\frac{x}{y}}=c\)

3 \(x y^{\frac{x}{y}} c\)

4 \(y+x e^{\frac{x}{y}}=c\)

Explanation:

(C) : Given,
\(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\)
\(\Rightarrow\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}=-\mathrm{e}^{\mathrm{x} / \mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}\)
\(\Rightarrow \frac{d x}{d y}=-\frac{e^{x / y}}{\left(1+e^{x / y}\right)}\left(1-\frac{x}{y}\right)\)
This is homogeneous differential equation.
So, put \(\mathrm{x}=\mathrm{vy}\)
\(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
Therefore, \(v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}\)
[From equation (i) and (ii)]
\(\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}-\mathrm{v}\)
\(=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}\left(1+\mathrm{e}^{\mathrm{v}}\right)}{1+\mathrm{e}^{\mathrm{v}}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}-\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}\)
\(y \frac{d v}{d y}=\frac{-\left(e^{v}+v\right)}{1+e^{v}}\)
\(\Rightarrow \frac{1+e^{v}}{v+e^{v}} d v=-\frac{d y}{y} \quad\) [Variable separation]
\(\Rightarrow \int \frac{1+e^{v}}{v+e^{v}} d v=-\int \frac{d y}{y}\)
Let \(\mathrm{v}+\mathrm{e}^{\mathrm{v}}=\mathrm{t} \Rightarrow\left(1+\mathrm{e}^{\mathrm{v}}\right) \mathrm{dy}=\mathrm{dt}\)
\(\therefore \int \frac{1+\mathrm{e}^{\mathrm{v}}}{\mathrm{v}+\mathrm{e}^{\mathrm{v}}} \mathrm{dv}=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log \mathrm{t}=\log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)\)
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=-\log \mathrm{y}+\log \mathrm{C} \quad\) [From equation (iii) and (iv)]
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)+\log \mathrm{y}=\log \mathrm{C}\)
\(\Rightarrow \mathrm{y}\left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=\mathrm{C}\)
\(\Rightarrow y\left(\frac{x}{y}+e^{x / y}\right)=C \Rightarrow y\left(\frac{x+y e^{x / y}}{y}\right)=C\)
\(\Rightarrow \mathrm{x}+\mathrm{ye}^{\mathrm{x} / \mathrm{y}}=\mathrm{C}\)

WB JEE-2019

Differential Equation

87324 General solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0\) is ( \(C\) is an arbitrary constant)

1 \(\frac{x}{a}=\tan \frac{y}{a}+c\)

2 \(\tan x y=c\)

3 \(\tan (x+y)=c\)

4 \(\tan \frac{y+c}{a}=\frac{x+y}{a}\)

Explanation:

(D) : Given,
Let,
\((x+y)^{2} \frac{d y}{d x}=a^{2}, a=0\)
et, \(\quad x+y=t\)
\(1+\frac{d y}{d x}=\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1\)
Therefore, \(\mathrm{t}^{2}\left(\frac{\mathrm{dt}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\)
\(\Rightarrow \quad \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{a}^{2}+\mathrm{t}^{2}\)
\(\frac{\mathrm{t}^{2}}{\mathrm{a}^{2}+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dx} \quad[\text { Variable separation] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\int \mathrm{dx} \quad \text { [int egration] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}+\mathrm{a}^{2}-\mathrm{a}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int\left(1-\frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}}\right) \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int \mathrm{dt}-\int \frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\mathrm{a}^{2}}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2}\left[\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right]=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow(\mathrm{x}+\mathrm{y})-\mathrm{a} \tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}\right)\)
\(=\mathrm{x}+\mathrm{C}^{\prime}[\) on putting the valueof \(\mathrm{t}]\)
\(\Rightarrow \frac{y-C^{\prime}}{a}=\tan ^{-1}\left(\frac{x+y}{a}\right)\)
\(\Rightarrow \frac{x+y}{a}=\tan \left(\frac{y-C^{\prime}}{a}\right) \Rightarrow \frac{x+y}{a}=\tan \left(\frac{y+C}{a}\right)\)
[where, \(\mathrm{C}=-\mathrm{C}^{\prime}\) is an arbitrary constant]
\(\tan \frac{\mathrm{y} C}{\mathrm{a}} \frac{\mathrm{x} y}{\mathrm{a}}\)

WB JEE-2019

Differential Equation

87325 The differential equation formed by eliminating \(a\) and \(b\) from the equation \(y=e^{x}(\operatorname{acos} x+b \sin x)\) is

1 \(2 \frac{d^{2} y}{d^{2}}+\frac{d y}{d x}-2 y=0\)

2 \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-2 y=0\)

3 \(2 \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}+2 y=0\)

4 \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\)

AP EAMCET-21.04.2019

Differential Equation

87327 The solution of the differential equation \(y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y\) satisfying \(y\left(\frac{\pi}{4}\right)=1\) is

1 \(\cos \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

2 \(\sin \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

3 \(\sin \frac{x}{y}=\log _{e} x-\frac{1}{\sqrt{2}}\)

4 \(\cos \frac{x}{y}=-\log _{e} x-\frac{1}{\sqrt{2}}\)

WB JEE-2013

Differential Equation

87323 The general solution of the differential equation \(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\) is \(\quad\) (c is an arbitrary constant)

1 \(x-y e^{\frac{x}{y}}=c\)

2 \(y-x e^{\frac{x}{y}}=c\)

3 \(x y^{\frac{x}{y}} c\)

4 \(y+x e^{\frac{x}{y}}=c\)

Explanation:

(C) : Given,
\(\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0\)
\(\Rightarrow\left(1+\mathrm{e}^{\mathrm{x} / \mathrm{y}}\right) \mathrm{dx}=-\mathrm{e}^{\mathrm{x} / \mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right) \mathrm{dy}\)
\(\Rightarrow \frac{d x}{d y}=-\frac{e^{x / y}}{\left(1+e^{x / y}\right)}\left(1-\frac{x}{y}\right)\)
This is homogeneous differential equation.
So, put \(\mathrm{x}=\mathrm{vy}\)
\(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
Therefore, \(v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}\)
[From equation (i) and (ii)]
\(\mathrm{y} \frac{\mathrm{dv}}{\mathrm{dy}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}-\mathrm{v}\)
\(=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}\left(1+\mathrm{e}^{\mathrm{v}}\right)}{1+\mathrm{e}^{\mathrm{v}}}=\frac{-\mathrm{e}^{\mathrm{v}}+\mathrm{ve}^{\mathrm{v}}-\mathrm{v}-\mathrm{ve}^{\mathrm{v}}}{1+\mathrm{e}^{\mathrm{v}}}\)
\(y \frac{d v}{d y}=\frac{-\left(e^{v}+v\right)}{1+e^{v}}\)
\(\Rightarrow \frac{1+e^{v}}{v+e^{v}} d v=-\frac{d y}{y} \quad\) [Variable separation]
\(\Rightarrow \int \frac{1+e^{v}}{v+e^{v}} d v=-\int \frac{d y}{y}\)
Let \(\mathrm{v}+\mathrm{e}^{\mathrm{v}}=\mathrm{t} \Rightarrow\left(1+\mathrm{e}^{\mathrm{v}}\right) \mathrm{dy}=\mathrm{dt}\)
\(\therefore \int \frac{1+\mathrm{e}^{\mathrm{v}}}{\mathrm{v}+\mathrm{e}^{\mathrm{v}}} \mathrm{dv}=\int \frac{\mathrm{dt}}{\mathrm{t}}=\log \mathrm{t}=\log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)\)
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=-\log \mathrm{y}+\log \mathrm{C} \quad\) [From equation (iii) and (iv)]
\(\Rightarrow \log \left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)+\log \mathrm{y}=\log \mathrm{C}\)
\(\Rightarrow \mathrm{y}\left(\mathrm{v}+\mathrm{e}^{\mathrm{v}}\right)=\mathrm{C}\)
\(\Rightarrow y\left(\frac{x}{y}+e^{x / y}\right)=C \Rightarrow y\left(\frac{x+y e^{x / y}}{y}\right)=C\)
\(\Rightarrow \mathrm{x}+\mathrm{ye}^{\mathrm{x} / \mathrm{y}}=\mathrm{C}\)

WB JEE-2019

Differential Equation

87324 General solution of \((x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0\) is ( \(C\) is an arbitrary constant)

1 \(\frac{x}{a}=\tan \frac{y}{a}+c\)

2 \(\tan x y=c\)

3 \(\tan (x+y)=c\)

4 \(\tan \frac{y+c}{a}=\frac{x+y}{a}\)

Explanation:

(D) : Given,
Let,
\((x+y)^{2} \frac{d y}{d x}=a^{2}, a=0\)
et, \(\quad x+y=t\)
\(1+\frac{d y}{d x}=\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1\)
Therefore, \(\mathrm{t}^{2}\left(\frac{\mathrm{dt}}{\mathrm{dx}}-1\right)=\mathrm{a}^{2}\)
\(\Rightarrow \quad \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{t}^{2}=\mathrm{a}^{2} \Rightarrow \mathrm{t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{a}^{2}+\mathrm{t}^{2}\)
\(\frac{\mathrm{t}^{2}}{\mathrm{a}^{2}+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dx} \quad[\text { Variable separation] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\int \mathrm{dx} \quad \text { [int egration] }\)
\(\Rightarrow \int \frac{\mathrm{t}^{2}+\mathrm{a}^{2}-\mathrm{a}^{2}}{\left(\mathrm{t}^{2}+\mathrm{a}^{2}\right)} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int\left(1-\frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}}\right) \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \int \mathrm{dt}-\int \frac{\mathrm{a}^{2}}{\mathrm{t}^{2}+\mathrm{a}^{2}} \mathrm{dt}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2} \int \frac{\mathrm{dt}}{\mathrm{t}^{2}+\mathrm{a}^{2}}=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow \mathrm{t}-\mathrm{a}^{2}\left[\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{t}}{\mathrm{a}}\right)\right]=\mathrm{x}+\mathrm{C}^{\prime}\)
\(\Rightarrow(\mathrm{x}+\mathrm{y})-\mathrm{a} \tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}\right)\)
\(=\mathrm{x}+\mathrm{C}^{\prime}[\) on putting the valueof \(\mathrm{t}]\)
\(\Rightarrow \frac{y-C^{\prime}}{a}=\tan ^{-1}\left(\frac{x+y}{a}\right)\)
\(\Rightarrow \frac{x+y}{a}=\tan \left(\frac{y-C^{\prime}}{a}\right) \Rightarrow \frac{x+y}{a}=\tan \left(\frac{y+C}{a}\right)\)
[where, \(\mathrm{C}=-\mathrm{C}^{\prime}\) is an arbitrary constant]
\(\tan \frac{\mathrm{y} C}{\mathrm{a}} \frac{\mathrm{x} y}{\mathrm{a}}\)

WB JEE-2019

Differential Equation

87325 The differential equation formed by eliminating \(a\) and \(b\) from the equation \(y=e^{x}(\operatorname{acos} x+b \sin x)\) is

1 \(2 \frac{d^{2} y}{d^{2}}+\frac{d y}{d x}-2 y=0\)

2 \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-2 y=0\)

3 \(2 \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}+2 y=0\)

4 \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\)

AP EAMCET-21.04.2019

Differential Equation

87327 The solution of the differential equation \(y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y\) satisfying \(y\left(\frac{\pi}{4}\right)=1\) is

1 \(\cos \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

2 \(\sin \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}\)

3 \(\sin \frac{x}{y}=\log _{e} x-\frac{1}{\sqrt{2}}\)

4 \(\cos \frac{x}{y}=-\log _{e} x-\frac{1}{\sqrt{2}}\)

WB JEE-2013

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Question Bank Series

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