Question #293047

 3. Find the particular solution of the linear equation dy/dx + 1/x y = x if y(1) = 0


1
Expert's answer
2022-02-02T15:15:33-0500
y+1xy=xy'+\dfrac{1}{x}y=x

Integration factor


μ(x)=e(dx/x)=x\mu(x)=e^{\int(dx/x)}=x

xy+y=x2xy'+y=x^2

d(xy)=x2dxd(xy)=x^2 dx

Integrate


d(xy)=x2dx\int d(xy)=\int x^2 dx

xy=x33+Cxy=\dfrac{x^3}{3}+C

y=x23+Cxy=\dfrac{x^2}{3}+\dfrac{C}{x}

Given y(1)=0y(1)=0


0=(1)23+C1=>C=130=\dfrac{(1)^2}{3}+\dfrac{C}{1}=>C=-\dfrac{1}{3}

The solution of the initial value problem is


y=x2313xy=\dfrac{x^2}{3}-\dfrac{1}{3x}


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