Question
Find the general solution of $x \frac{d y}{d x}+2 y=x^{2} \log x$
✓
Answer
It is given that $x \frac{d y}{d x}+2 y=x^{2} \log x$
$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = $\frac{2}{x}$ and Q = x log x)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{2(\log \mathrm{x})}=\mathrm{e}^{\log \mathrm{x}^{2}}=\mathrm{x}^{2}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} )=\int(\mathrm{Q} \times \mathrm{I.F}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{x}^{2}=\int\left(\mathrm{x} \log \mathrm{x} \cdot \mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow x^{2} y=\int\left(x^{3} \log x\right) d x+c$
$\Rightarrow x^{2} y=\log x . \int x^{3} d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^{3} d x\right] d x+c$
$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int\left(\frac{1}{x} \cdot \frac{x^{4}}{4}\right) d x+c$
$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+C$
$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+c$
$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+c$
$\Rightarrow \mathrm{y}=\frac{1}{16} \mathrm{x}^{2}(4 \log \mathrm{x}-1)+\mathrm{Cx}^{-2}$
Therefore, the required general solution of the given differential equation
$y=\frac{1}{16} x^{2}(4 \log x-1)+C x^{-2}$
$\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = $\frac{2}{x}$ and Q = x log x)
Now, I.F. = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{2(\log \mathrm{x})}=\mathrm{e}^{\log \mathrm{x}^{2}}=\mathrm{x}^{2}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} )=\int(\mathrm{Q} \times \mathrm{I.F}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{y} \cdot \mathrm{x}^{2}=\int\left(\mathrm{x} \log \mathrm{x} \cdot \mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow x^{2} y=\int\left(x^{3} \log x\right) d x+c$
$\Rightarrow x^{2} y=\log x . \int x^{3} d x-\int\left[\frac{d}{d x}(\log x) \cdot \int x^{3} d x\right] d x+c$
$\Rightarrow x^{2} y=\log x \cdot \frac{x^{4}}{4}-\int\left(\frac{1}{x} \cdot \frac{x^{4}}{4}\right) d x+c$
$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \int x^{3} d x+C$
$\Rightarrow x^{2} y=\frac{x^{4} \log x}{4}-\frac{1}{4} \cdot \frac{x^{4}}{4}+c$
$\Rightarrow x^{2} y=\frac{1}{16} x^{4}(4 \log x-1)+c$
$\Rightarrow \mathrm{y}=\frac{1}{16} \mathrm{x}^{2}(4 \log \mathrm{x}-1)+\mathrm{Cx}^{-2}$
Therefore, the required general solution of the given differential equation
$y=\frac{1}{16} x^{2}(4 \log x-1)+C x^{-2}$
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