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Exercises · Q15

Q.Solve the linear differential equation dydx+y=x\displaystyle\frac{dy}{dx} + y = x.

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Standard linear form with P=1P = 1 and Q=xQ = x.

Integrating factor: ∫P dx=∫1 dx=x\int P\,dx = \int 1\,dx = x, so I.F.=ex\text{I.F.} = e^{x}.

General solution:

y⋅ex=∫x ex dx+c.y\cdot e^{x} = \int x\,e^{x}\,dx + c.

Integrate ∫xex dx\int x e^{x}\,dx by parts (with u=xu=x, dv=exdxdv=e^{x}dx): xex−∫ex dx=xex−ex=ex(x−1)x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} = e^{x}(x-1). Hence

y ex=ex(x−1)+c.y\,e^{x} = e^{x}(x - 1) + c.

Divide by exe^{x}:

y=x−1+c e−x.y = x - 1 + c\,e^{-x}. …

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