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

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(A) We have,$\frac{dy}{dx} - y = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$. The i

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We have,$\frac{dy}{dx} - y = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$. The integrating factor is $IF = e^{\int P dx} = e^{\int -1 dx} = e^{-x}$. The general solution is given by $y \cdot IF = \int Q \cdot IF dx + c$. Substituting the values,we get $y e^{-x} = \int x^2 e^{-x} dx + c$. Using integration by parts,$\int x^2 e^{-x} dx = -x^2 e^{-x} + \int 2x e^{-x} dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + c$. Thus,$y e^{-x} = -e^{-x}(x^2 + 2x + 2) + c$. Multiplying both sides by $e^x$,we get $y = -(x^2 + 2x + 2) + ce^x$. Rearranging the terms,we obtain $y + x^2 + 2x + 2 = ce^x$.

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

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