Question

Resolver a seguinte equação diferencial pelo método de variação de parâmetro:

a) y′′ − y4′ + y4 = ( x + 1)е²ˣ

Answer 1

Resposta:

$y(x)=C_{1}e^{2x}+C_{2}xe^{2x}+\frac{x^3}{6}e^{2x}+\frac{x^2}{2}e^{2x}$

Explicação passo-a-passo:

$w^2-4w+4=0\\(w-2)^2=0\\w=2\\y_{h}=C_{1}e^{2x}+C_{2}xe^{2x}\\\\y_{p}=a(x)e^{2x}+b(x)xe^{2x}\\\left \{{{a^{'}(x)e^{2x}+b^{'}(x)xe^{2x}=0} \atop {2a^{'}(x)e^{2x}+b^{'}(x)e^{2x}(2x+1)=e^{2x}(x+1)}} \right\\\left \{ {{a^{'}(x)+b^{'}(x)x=0} \atop {2a^{'}(x)+b^{'}(x)(2x+1)=x+1}} \right\\a^{'}(x)=-x^2-x\\b^{'}(x)=x+1\\y_{p}=\frac{x^3e^{2x}}{6}+\frac{x^2e^{2x}}{2}\\\\y(x)=y_{h}+y_{p}\\y(x)=C_{1}e^{2x}+C_{2}xe^{2x}+\frac{x^3}{6}e^{2x}+\frac{x^2}{2}e^{2x}$