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Question

(A) The given differential equation is: $\frac{dy}{dx} + y \cot x = 4x \csc x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py

Vedclass · Accepted Answer

$y \sin x = 2x^2 - \frac{\pi^2}{2}$

Vedclass · Answer

(A) The given differential equation is: $\frac{dy}{dx} + y \cot x = 4x \csc x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 4x \csc x$. First,we find the integrating factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$. The general solution is given by $y(I.F.) = \int (Q \cdot I.F.) dx + C$. Substituting the values: $y \sin x = \int (4x \csc x \cdot \sin x) dx + C$. Since $\csc x \cdot \sin x = 1$,we have $y \sin x = \int 4x dx + C$. Integrating,we get $y \sin x = 2x^2 + C$. Given that $y=0$ when $x=\frac{\pi}{2}$,we substitute these values: $0 \cdot \sin(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 + C$. $0 = 2(\frac{\pi^2}{4}) + C \Rightarrow 0 = \frac{\pi^2}{2} + C \Rightarrow C = -\frac{\pi^2}{2}$. Wait,re-evaluating the constant: $0 = 2(\frac{\pi^2}{4}) + C \Rightarrow 0 = \frac{\pi^2}{2} + C \Rightarrow C = -\frac{\pi^2}{2}$. Thus,the particular solution is $y \sin x = 2x^2 - \frac{\pi^2}{2}$.

Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \csc x$ $(x \neq 0),$ given that $y=0$ when $x=\frac{\pi}{2}$.

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