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Question

(B) Given the differential equation: $x \frac{dy}{dx} + y = x e^x$.Dividing by $x$ (assuming $x 
eq 0$), we get: $\frac{dy}{dx} + \frac{1}{x} y = e^x

Vedclass · Accepted Answer

$x-1$

Vedclass · Answer

(B) Given the differential equation: $x \frac{dy}{dx} + y = x e^x$.Dividing by $x$ (assuming $x 
eq 0$), we get: $\frac{dy}{dx} + \frac{1}{x} y = e^x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = e^x$.The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.The solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + C$.$xy = \int x e^x dx + C$.Using integration by parts for $\int x e^x dx$:$\int x e^x dx = x e^x - \int 1 \cdot e^x dx = x e^x - e^x = e^x(x-1)$.Thus, $xy = e^x(x-1) + C$.Comparing this with the given form $xy = e^x \phi(x) + C$, we find $\phi(x) = x-1$.

If the solution of the differential equation $x \frac{dy}{dx} + y = x e^x$ is $xy = e^x \phi(x) + C$, then $\phi(x)$ is equal to

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