y + x^2 = frac{dy}{dx} has the solution | vedclass

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

Vedclass · Accepted Answer

$y + x^2 + 2x + 2 = ce^x$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} - y = x^2$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$.The Integrating Factor $(I.F.)$ is given by $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.Substituting the values,$y e^{-x} = \int x^2 e^{-x} dx + c$.Using integration by parts,$\int x^2 e^{-x} dx = x^2(-e^{-x}) - \int 2x(-e^{-x}) dx = -x^2 e^{-x} + 2 \int x e^{-x} dx$.$= -x^2 e^{-x} + 2[x(-e^{-x}) - \int 1(-e^{-x}) dx] = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + c$.Thus,$y e^{-x} = -e^{-x}(x^2 + 2x + 2) + c$.Multiplying both sides by $e^x$,we get $y = -(x^2 + 2x + 2) + ce^x$,which simplifies to $y + x^2 + 2x + 2 = ce^x$.

$y + x^2 = \frac{dy}{dx}$ has the solution

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