Answer:
[tex]y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}[/tex].
Step-by-step explanation:
Using first order linear differential equation:
[tex]\dfrac{\mathrm{d} y}{\mathrm{d} t} = 2y + t^2[/tex]
[tex]\frac{\mathrm{d} y}{\mathrm{d} t} - 2y =t^2[/tex]
finding integrating factor:
I.F = [tex] e^{\int -2dt}[/tex]
I.F =[tex]e^{-2t}[/tex]
now,
[tex]y = \dfrac{1}{IF}(\int t^2dt+ c )[/tex]
[tex]y = \dfrac{1}{e^{-2t}}(\int t^2dt+ c )[/tex]
[tex]y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}[/tex]
hence the solution is
[tex]y = \dfrac{t^3e^{2t}}{3}+Ce^{2t}[/tex]
