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Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 7

Answers (1)

Answer:

 \Rightarrow xy=\frac{x^{2}}{2}+c

Hint:

 Using the form

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

Given:

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

Solution:

Comparing with the given equation with

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

We get,

P=\frac{1}{x} \text { and }Q=1

I.F=e^{\int P\, dx}=e^{\int\frac{1}{x} dx}=e^{log\, x}=x

∴The general solution is

\begin{aligned} &y\times I.F=\int Q\times I.F\, dx \\ &\Rightarrow y\times x=\int 1\times x\, dx \\ &\Rightarrow xy=\frac{x^{2}}{2}+c \end{aligned}

So the answer is

\begin{aligned} &\Rightarrow xy=\frac{x^{2}}{2}+c \end{aligned}

Posted by

Gurleen Kaur

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