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If \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{xy}{x^{2}+y^{2}};y(1)=1; then a value of x satisfying y(x)=e is :   
Option: 1 \sqrt{3}\: e
 
Option: 2 \frac{1}{2}\sqrt{3}\: e
 
Option: 3 \sqrt{2}\: e
 
Option: 4 \frac{e}{\sqrt{2}}
 
 

Answers (4)

best_answer

 

 

Homogeneous Differential Equation -

This equation can be solved by the substitution y = vx.

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;y=vx}\\\mathrm{\Rightarrow\;\;\;\;\;\;\;\;\;\;\; \frac{dy}{dx}=v+x\frac{dv}{dx}}\\\text { Thus, } \frac{d y}{d x}=\phi\left(\frac{y}{x}\right) \text { transforms to }\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; v+x\frac{dv}{dx}=\phi\left ( v \right )}\\\\\Rightarrow\;\;\;\;\;\;\;\;\;\;\; \frac{d v}{\phi(v)-v}=\frac{d x}{x}\\\text{The variables have now been separated and the solution is}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;}\int \frac{\mathrm{d} \mathrm{v}}{\phi(\mathrm{v})-\mathrm{v}}=\ln \mathrm{x}+\mathrm{c}\\\\\text{After the integration v should be replaced by y/x} \\\text{to get the required solution.}

 

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\text { put } y=v x, \quad \frac{d y}{d x}=v+x \frac{d v}{d x}

v+x \frac{d v}{d x}=\frac{ v}{1+v^{2}}

x \frac{d v}{d x}=\frac{-v^{3}}{1+v^{2}}

\frac{1+v^{2}}{v^{3}} d v=-\frac{d x}{x}

-\frac{x^{2}}{2 y^{2}}+\ln y=c \Rightarrow c=-1 / 2

-\frac{x^{2}}{2 y^{2}}+\ln y=c \Rightarrow c=-1 / 2

\text { Now } y(x)=e \quad \Rightarrow \quad 1=\frac{x^{2}}{2 e^{2}}-\frac{1}{2}

x=\pm \sqrt{3} e

Correct Option 1

Posted by

avinash.dongre

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Option: 3

Posted by

Ayesha Sabeela

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C5

X-45

 

Posted by

Aman thakur

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√3e

Posted by

Nalla mahalakshmi

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