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

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

Answers (1)

2021-03-10
\(dy/dx+P(x)*y= Q(x)\)
\(I.F.=e(int Pdx)\)
\(y(I.F.)= int Qx(I.F.)dx+C\)
\(P(x)= a/x\)
\(Q(x)= 40x\)
\(I.F.= e(int a/x dx)= e^(a lnx)=e^(ln x^a)=x^a\)
\(y(I.F.)= int(40x)(I.F.)dx + C\)
\(y(x^a)= int(40x)(x^a)dx + C\)
\(y x^a= 40 int x^(a+1)dx+C\)
\(y x^a= 40*(x^2+1+1)/(a+1+1)+C\)
\(y x^a= 40*(x^(a+2)/(a+2))+C\)
\(y= 40/(a+2)*x^(a+2)/x^a+C/x^a\)
\(y= 40/(a+2)*x^2+C/x^a\) (1)
y(1)=a
Put x=1 and y=a into equation (1)
\(a= 40/(a+2)(1)^2+C/1^a\)
\(a= 40/(a+2)+C\)
\(C= a-40/(a+2)\)
\(C= (a(a+2)-40)/((a+2))\)
\(C= (a^2+2a-40)/(a+2)\) Put the value of C into equation (1) \(y= 40/(a+2) x^2+(a^2+2a-40)/(a+2)* 1/x^a\)
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