Find the general solution for each of the following differential equations. $\left(1-x^{2}\right) \frac{d y}{d x}+x y=a x$

CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal

Solution:

$\left(1-x^{2}\right) \frac{d y}{d x}=x y=a x-\ldots(1)$
To solve (1) we will use following formula
$\begin{array}{r} \int \frac{f^{\prime}(X)}{f(x)} d x=\log f(x)+\mathrm{C} \\ a \log b=\log b^{a} \\ a^{\log _{a} b}=\log b \end{array}$
General solution for the differential equation in the form of is given by,
$\begin{array}{c} \frac{d y}{d x}+P y=Q \\ y \cdot(I . F .)=\int Q \cdot(\text { I.F. }) d x+c \end{array}$
Where, integrating factor,
$\text { I. } F .=e \int^{p d x}$
Dividing equation (1) by $\left(1-x^{2}\right)$ ,
$\frac{d y}{d x}+\frac{x}{\left(1-x^{2}\right)} \cdot y=\frac{a x}{\left(1-x^{2}\right)} \ldots \ldots \ldots(2)$
Comparing (2) with
$\frac{d y}{d x}+P y=Q$
Where, $P=\frac{x}{\left(1-x^{2}\right)}$ and $Q=\frac{a x}{\left(1-x^{2}\right)}$
Therefore, integrating factor is
$\begin{aligned} \text { I. } F .=e \int p d x \\ = e^{\int \frac{x}{\left(1-x^{2}\right)} d x} \\ = e^{\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)} d x} \end{aligned}$
$\operatorname{Let}\left(1-x^{2}\right)=f(x)$
Thus $f^{\prime}(x)=-2 x$
$\begin{array}{c} \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{-2 x}{\left(1-x^{2}\right)} d x=\log f(x)=\log \left(1-x^{2}\right) \\ \text { I.F. }=e^{\frac{-1}{2} \log \left(1-x^{2}\right)} \\ e^{\log \left(1-x^{2}\right)^{-1 / 2}}\left(a \log b=\log b^{a}\right) \\ \qquad e^{\log \left(\frac{1}{\sqrt{1-x^{2}}}\right)} \\ \frac{1}{\sqrt{1-x^{2}}}-\left(a^{\log _{a} b}=\log b\right) \end{array}$
General solution is
$\begin{array}{c} y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c \\ y \cdot\left(\frac{1}{\sqrt{1-x^{2}}}\right)=\int\left(\frac{a x}{\sqrt{1-x^{2}}}\right) \cdot\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x+c \end{array}$
Let
$I=\int \frac{a x}{\left(1-x^{2}\right)^{3 / 2}} d x$
$\operatorname{Put}\left(1-\mathrm{x}^{2}\right)=\mathrm{t}$
$\begin{array}{r} -2 x d x=d t \\ x d x=\frac{-d t}{2} \\ I=\int \frac{a}{t^{3 / 2}} \cdot \frac{-d t}{2} \\ I=\frac{-a}{2} \int t^{3 / 2} d t \\ I=\frac{-a}{2} \cdot \frac{t^{-1 / 2}}{-1 / 2} \\ I=a \cdot \frac{1}{\sqrt{t}} \\ I=\frac{a}{\sqrt{1-x^{2}}} \end{array}$
Substituting I in (3)
$\frac{y}{\sqrt{1-x^{2}}}=\frac{a}{\sqrt{1-x^{2}}}+c$
Multiplying above equation by $\sqrt{1-x^{2}}$
$y=a+c \sqrt{1-x^{2}}$