Solve the following differential equations:cosec x dy dx +x^2y^2=… - Vidyadip

Question

Solve the following differential equations:$\text{cosec x}\log\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^2\text{y}^2=0$

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Answer

$\text{cosec x}\log\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^2\text{y}^2=0$
$\Rightarrow\text{cosec x }\log\text{y}\frac{\text{dy}}{\text{dx}}=-\text{x}^2\text{y}^2$
$\Rightarrow\frac{1}{\text{y}^2}\log\text{ y dy}=-\frac{\text{x}^2}{\text{cosec x}}\text{dx}$
$\Rightarrow\frac{1}{\text{y}^2}\log\text{ y dy}=-\text{x}^2\sin\text{x dx}$
$\Rightarrow\int\frac{1}{\text{y}^2}\log\text{ y dy}=-\int\text{x}^2\sin\text{x dx}$
$\Rightarrow-\frac{\log\text{y}}{\text{y}}+\int\frac{1}{\text{y}}\times\frac{1}{\text{y}}=-\Big[-\text{x}^2\cos\text{x}+\int2\text{x}\cos\text{x dx}\Big]+\text{C}$
$\Rightarrow-\frac{\log\text{y}}{\text{y}}-\frac{1}{\text{y}}=-\Big[-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}-2\int\sin\text{x dx}\Big]+\text{C}$
$\Rightarrow-\Big(\frac{1+\log\text{y}}{\text{y}}\Big)=-\big[-\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}+2\cos\text{x dx}\big]+\text{C}$
$\Rightarrow-\Big(\frac{1+\log\text{y}}{\text{y}}\Big)-\text{x}^2\cos\text{x}+2(\text{x}\sin\text{x}+\cos\text{x})=\text{C}$

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