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Find the general solution for each of the following differential equations. \left(1-x^{2}\right) \frac{d y}{d x}+x y=x \sqrt{1-x^{2}}
CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal
Find the general solution for each of the following differential equations. \left(1-x^{2}\right) \frac{d y}{d x}+x y=x \sqrt{1-x^{2}}

Solution:

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

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