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

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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