Question

Solve differential equation dy/dx+(a/x)y=40x, for x>0 and y(1)=a

First order differential equations
ANSWERED
asked 2021-03-09

Solve differential equation \(\frac{dy}{dx}+(\frac{a}{x})y=40x\), for \(x>0\) and \(y(1)=a\)

Expert Answers (1)

2021-03-10

\(\frac{dy}{dx}+P(x)\times y= Q(x)\)
\(I.F.=e(\int Pdx)\)
\(y(I.F.)= \int Qx(I.F.)dx+C\)
\(P(x)= \frac{a}{x}\)
\(Q(x)= 40x\)
\(I.F.= e(\int \frac{a}{x} dx)= e^{a \ln x}=e^{\ln x^a}=x^a\)
\(y(I.F.)= \int(40x)(I.F.)dx + C\)
\(y(x^a)= \int(40x)(x^a)dx + C\)
\(y x^a= 40 \int x^{a+1}dx+C\)
\(y x^a= 40\times \frac{x^2+1+1}{a+1+1}+C\)
\(y x^a= 40\times\frac{x^{a+2}}{a+2}+C\)
\(y= \frac{40}{a+2}\times \frac{x^{a+2}}{x^a}+\frac{C}{x^a}\)
\(y= \frac{40}{a+2}\times x^2+\frac{C}{x^a}\) (1)
\(y(1)=a\)
Put \(x=1\) and \(y=a\) into equation (1)
\(a= \frac{40}{a+2}(1)^2+\frac{C}{1^a}\)
\(a= \frac{40}{a+2}+C\)
\(C= a-\frac{40}{a+2}\)
\(C= \frac{a(a+2)-40}{a+2}\)
\(C= \frac{a^2+2a-40}{a+2}\) Put the value of C into equation (1) \(y= \frac{40}{a+2} x^2+\frac{a^2+2a-40}{a+2}\times \frac{1}{x^a}\)

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