The solution of the differential equation x f | vedclass

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Question

(C) Given the differential equation: $x \frac{dy}{dx} = y + x^2$. Dividing both sides by $x$ (assuming $x 
eq 0$): $\frac{dy}{dx} - \frac{1}{x} y = x

Vedclass · Accepted Answer

$y = x^2 + ax$

Vedclass · Answer

(C) Given the differential equation: $x \frac{dy}{dx} = y + x^2$. Dividing both sides by $x$ (assuming $x 
eq 0$): $\frac{dy}{dx} - \frac{1}{x} y = x$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = x$. The integrating factor $(I.F.)$ is given by $e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\log_e x} = e^{\log_e (x^{-1})} = \frac{1}{x}$. The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + a$. Substituting the values: $y \cdot \frac{1}{x} = \int x \cdot \frac{1}{x} dx + a$. $\frac{y}{x} = \int 1 dx + a$. $\frac{y}{x} = x + a$. Therefore,$y = x^2 + ax$.

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

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