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Please solve RD Sharma class 12 chapter 21 Differential Equation exercise Fill in the blank question 17 maths textbook solution

Answers (1)

Answer:

 \frac{e^{x}}{x}

Hint:

 Use the form

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q,

to find the integrating factor.

Given:

 \frac{\mathrm{d} y}{\mathrm{d} x}+y=\frac{1+y}{x}

Solution:

\frac{\mathrm{d} y}{\mathrm{d} x}+y-\frac{y}{x}=\frac{1}{x}

\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+y\left ( 1-\frac{1}{x} \right )=\frac{1}{x}

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q,

Where

P=\left ( 1-\frac{1}{x} \right )

I.F=e^{\int P\, dx}=e^{\int(1-\frac{1}{x}) dx}=e^{(x-in\, x)}

I.F=e^{x}.e^{-in\, x}=e^{x}.e^{in_{e}\, (x^{-1})}=e^{x}.\frac{1}{x}=\frac{e^{x}}{x}

So, the answer is

\frac{e^{x}}{x}

Posted by

Gurleen Kaur

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