Answer:
[tex]y=\frac{t^2e^{2t}}{3}+ce^{2t}[/tex]
Step-by-step explanation:
We have given differential equation [tex]\frac{dy}{dt}-2y=t^2e^{2t}[/tex]
We know that linear differential equation is given by [tex]\frac{dy}{dt}+Py=Q[/tex]
On comparing with standard equation P = -2 and Q= [tex]t^2e^{2t}[/tex]
Now integrating factor [tex]IF=e^{-Pdt}[/tex]
[tex]IF=e^{-2dt}=e^{-2t}[/tex]
Now solution of differential equation is given by
[tex]y\times IF=\int\ IF\times Q\ dt[/tex]
[tex]y\times e^{-2t}=\int\ e^{-2t}\times t^2e^{2t}\ dt[/tex]
[tex]y\times e^{-2t}=\frac{t^2}{3}+c[/tex]
[tex]y=\frac{t^2e^{2t}}{3}+ce^{2t}[/tex]