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Question

(A) The given differential equation is $\frac{dy}{dx} + (2 \cot x)y = 3x^2 \csc^2 x$.This is a linear differential equation of the form $\frac{dy}{dx}

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$y \sin^2 x = x^3 + c$

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(A) The given differential equation is $\frac{dy}{dx} + (2 \cot x)y = 3x^2 \csc^2 x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 \cot x$ and $Q = 3x^2 \csc^2 x$.First,we find the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 2 \cot x dx} = e^{2 \ln|\sin x|} = e^{\ln(\sin^2 x)} = \sin^2 x$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values:$y \cdot \sin^2 x = \int (3x^2 \csc^2 x) \cdot \sin^2 x dx$.Since $\csc^2 x = \frac{1}{\sin^2 x}$,the expression simplifies to:$y \sin^2 x = \int 3x^2 dx$.Integrating $3x^2$ with respect to $x$ gives $x^3$.Thus,$y \sin^2 x = x^3 + c$.

The solution of the differential equation $\frac{dy}{dx} + 2y \cot x = 3x^2 \csc^2 x$ is

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