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Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)5 Marks
Question
Solve $y d x-x d y+\log x d x=0$
✓

Answer

$
\begin{aligned}
& y d x-x d y+\log x d x=0 \\
& \therefore(y+\log x) d x=x d y \\
& \therefore \frac{y+\log x}{x}=\frac{d y}{d x} \\
& \therefore \frac{y}{x}+\frac{\log x}{x}=\frac{d y}{d x} \\
& \therefore \frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x}
\end{aligned}
$
This is a linear differential equation of the form
$
\frac{d y}{d x}+P y=Q \text {, where } P=-\frac{1}{x}, Q=\frac{\log x}{x}
$
$
\begin{aligned}
\therefore \text { I.F. } & =e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\int \frac{1}{x} d x} \\
& =e^{-\log x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x}
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
y \cdot\left(\text { I.F.) }=\int Q \cdot \text { (I.F.) } d x+c\right.
$
$
\begin{aligned}
\therefore \frac{y}{x} & =\int \frac{\log x}{x} \cdot \frac{1}{x} d x+c \\
& =\int \log x \cdot x^{-2} d x+c
\end{aligned}
$
$
\begin{aligned}
& =(\log x) \int x^{-2} d x-\int\left[\left\{\frac{d}{d x}(\log x)\right\} \int x^{-2} d x\right] d x \\
& =(\log x)\left(\frac{x^{-1}}{-1}\right)-\int \frac{1}{x} \cdot\left(\frac{x^{-1}}{-1}\right) d x+c
\end{aligned}
$
$
=-\frac{\log x}{x}+\int x^{-2} d x+c
$
$
\begin{aligned}
& \therefore \frac{y}{x}=-\frac{\log x}{x}-\frac{1}{x}+c \\
& \therefore y=c x-(1+\log x)
\end{aligned}
$

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