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Question

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

Vedclass · Accepted Answer

$4xy = x^4 + c$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = x^2$. First,we find the integrating factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is given by $y 	imes (I.F.) = \int (Q 	imes I.F.) dx + c$. Substituting the values,we get $y 	imes x = \int (x^2 	imes x) dx + c$. $xy = \int x^3 dx + c$. $xy = \frac{x^4}{4} + c$. Multiplying both sides by $4$,we get $4xy = x^4 + 4c$. Since $4c$ is an arbitrary constant,we can write it as $c$. Thus,the solution is $4xy = x^4 + c$.

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

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