The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation: $x^{2} dy - 2xy dx = x^{4} \cos x dx$. Dividing both sides by $x^{2} dx$ (assuming $x 
eq 0$),we get: $\frac{dy}

Vedclass · Accepted Answer

$y = x^{2} \sin x + cx^{2}$

Vedclass · Answer

(A) Given the differential equation: $x^{2} dy - 2xy dx = x^{4} \cos x dx$. Dividing both sides by $x^{2} dx$ (assuming $x 
eq 0$),we get: $\frac{dy}{dx} - \frac{2}{x}y = x^{2} \cos x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{2}{x}$ and $Q = x^{2} \cos x$. The integrating factor $(IF)$ is given by: $IF = e^{\int P dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln |x|} = e^{\ln |x^{-2}|} = \frac{1}{x^{2}}$. The general solution is $y \cdot IF = \int (Q \cdot IF) dx + c$. Substituting the values: $y \cdot \frac{1}{x^{2}} = \int (x^{2} \cos x) \cdot \frac{1}{x^{2}} dx + c$. $\frac{y}{x^{2}} = \int \cos x dx + c$. $\frac{y}{x^{2}} = \sin x + c$. Multiplying by $x^{2}$,we get the general solution: $y = x^{2} \sin x + cx^{2}$.

The general solution of the differential equation $x^{2} dy - 2xy dx = x^{4} \cos x dx$ is

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