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Please solve RD Sharma class 12 chapter 21 Differential Equation exercise Fill in the blank question 25 maths textbook solution

Answers (1)


 Integrating factor = x2 + 1


 we will use the linear differential equation to solve the problem.


 The integrating factor of all differential equation

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

is _________

Dividing the given equation both sides by x2 +1


Dividing the L.H.S by (x2 + 1) we get,

\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2x}{x^{2}+1}y=\frac{x^{2}-1}{x^{2}+1} \qquad \qquad \dots(i)

Clearly, the equation (i) is of the form

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q \qquad \qquad \dots(ii)

Comparing (i) and (ii) we get,

\begin{aligned} &P(x)=\frac{2x}{x^{2}+1} \\ &Q(x)=\frac{x^{2}-1}{x^{2}+1} \end{aligned}


\begin{aligned} &I.F.=e^{\int P(x)dx}=e^{\int \frac{2x}{x^{2}+1}dx} \end{aligned}

Now Put x2 +1 = t ⇒2x dx =dt

\begin{aligned} &I.F.=e^{\int \frac{2x}{x^{2}+1}dx} \\ &\Rightarrow e^{\int \frac{1}{t}dt} \\ &\Rightarrow e^{log\, t}= t \\ &\therefore t=x^{2}+1 \end{aligned}

So, the answer is x2 +1

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Gurleen Kaur

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