Question

Resolva os seguintes problemas de valor inicial.$\left \{ \frac{dy}{dx} =x . e^{2x} onde , y =(0)=1$

Answer 1

dy/dx = x*e^(2x)  

dy/dx = x*e^(2x)  

dy = x*e^(2x) dx

∫ dy = ∫ x*e^(2x) dx

Fazendo por partes

∫ x*e^(2x) dx

u=x ==> du=dx

dv = e^(2x) dx ==> ∫  dv = ∫  e^(2x) dx ==> v= (1/2)* e^(2x)

∫ x*e^(2x) dx =(x/2)* e^(2x) - ∫(1/2)* e^(2x) dx

∫ x*e^(2x) dx =(x/2)* e^(2x) -(1/2) ∫e^(2x) dx

∫ x*e^(2x) dx =(x/2)* e^(2x) -(1/2) *(1/2)*e^(2x) +c

∫ x*e^(2x) dx =(x/2)* e^(2x) -(1/4)*e^(2x) +c

y = (x/2)* e^(2x) -(1/4)*e^(2x) +c

para y(0)=1  

1 = (0/2)* e^(2*0) -(1/4)*e^(2*0) +c

1 = -(1/4)*e⁰ +c  ==>c=  1/4

y(x) = (x/2)* e^(2x) -(1/4)*e^(2x) +  1/4

y(x)=e^(2x) *[x/2 -1/4]  + 1/4