The solution of the differential equation   \frac{dy}{dx}=\frac{x+y}{x}    satisfying the condition y(1)=1  is

  • Option 1)

    y=x\; \ln \; x+x

  • Option 2)

    y= \ln \; x+x

  • Option 3)

    y=x\; \ln \; x+x^{2}

  • Option 4)

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

 

Answers (1)

As we learnt in 

 

Linear Differential Equation -

\frac{dy}{dx}+Py= Q

- wherein

P, Q are functions of x alone.

 

 \frac{dy}{dx}=\frac{x+y}{x}=1+\frac{y}{x}

\Rightarrow \frac{dy}{dx}-\frac{y}{x}=1

Put    P=-\frac{1}{x}\:\:,\:\:Q=1

\int -\frac{1}{x}\:dx=-\log x=\log \frac{1}{x}

I.F.=e^{log^{1/x}}=\frac{1}{x} 

Solution is y\:.\:\frac{1}{x}=\int \frac{1}{x}dx=\log x +C

\frac{1}{1}=\log 1+C\:\:\:\Rightarrow C=1

\therefore y=x+ \log x


Option 1)

y=x\; \ln \; x+x

This option is correct.

Option 2)

y= \ln \; x+x

This option is incorrect.

Option 3)

y=x\; \ln \; x+x^{2}

This option is incorrect.

Option 4)

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

This option is incorrect.

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