9. differential equations — ગણિત ધોરણ 12 સાયન્સ — Question

$\sec ^{2} y \frac{d y}{d x}+\frac{1}{x} 2 \tan y=x^{3}$

Put tan $y=z \Rightarrow \sec ^{2} y \frac{d y}{d x}=\frac{d z}{d x}$

$\Rightarrow \quad \frac{d z}{d x}+\frac{2}{x} z=x^{3},$ which is linear in $z$

The integrating factor is

I.F. $=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}$

Hence, the solution is

$\mathrm{z}\left(\mathrm{x}^{2}\right)=\int \mathrm{x}^{3}\left(\mathrm{x}^{2}\right) \mathrm{d} \mathrm{x}+\mathrm{a} \Rightarrow \mathrm{zx}^{2}=\frac{\mathrm{x}^{6}}{6}+\mathrm{a}$

$\therefore(6 \tan y) x^{2}=x^{6}+6 a$

$\Rightarrow 6 x^{2} \tan y=x^{6}+c,[c=6 a]$