Find the solution of the given initial value problem:(a) y' + 2y = te^{-2t}, y(1) = 0(b) t^{3}y' + 4t^{2}y = e^{-t}, y(-1) = 0

Answer:[tex](a)\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex][tex](b)\ y(t)=\ (1-t)e^{-t}\ -\ 2e[/tex]Step-by-step explanation:(a) [tex]y'\ +\ 2y\ =\ te^{-2t},\ y(1)\ =\ 0[/tex]  [tex]=>\ (D+2)y\ =\ te^{-2t}[/tex]To find the complementary function    D+2 = 0 => D = -2So, the complementary function can by given by[tex]y_c(t)\ =\ C.e^{-2t}[/tex]Now, to find particular integral   [tex](D+2)y_p(t)\ =\ te^{-2t}[/tex][tex]=>y_p(t)\ =\ \dfrac{ te^{-2t}}{D+2}[/tex]               [tex]=\ \dfrac{ te^{-2t}}{-2+2}[/tex]                = not definedSo, [tex]y_p(t)\ =\ \dfrac{ t^2e^{-2t}}{D^2}[/tex]            [tex]=\ \dfrac{t^2e^{-2t}}{(-2)^2}[/tex]            [tex]=\ \dfrac{t^2e^{-2t}}{4}[/tex]So, complete solution can be given by     [tex]y(t)\ =\ y_c(t)\ +\ y_p(t)[/tex][tex]=> y(t) =\ C.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex]As given in question[tex]=>\ y(1)\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex][tex]=>\ 0\ =\ C.e^{-2}\ +\ \dfrac{1^2e^{-2}}{4}[/tex][tex]=>\ C\ =\ 4e^2[/tex]Hence, the complete solution can be give by[tex]=>\ y(t) =\ 4e^2.e^{-2t}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex][tex]=>\ y(t) =\ 4.e^{2(1-t)}\ +\ \dfrac{t^2e^{-2t}}{4}[/tex](b) [tex]t^{3}y'\ +\ 4t^{2}y\ =\ e^{-t},\ y(-1)\ =\ 0[/tex][tex]=>\ y'\ +\ 4t^{-1}y\ =\ t^{-3}e^{-t}[/tex]Integrating factor can be given by[tex]I.F\ =\ e^{\int (4t^{-1})dt}[/tex]      [tex]=\ e^{log\ t^4}[/tex]      [tex]=\ t^4[/tex]Now , the solution of the given differential equation can be given by[tex]y(t)\times t^4\ =\ \int t^{-3}e^{-t}t^4dt\ +\ C[/tex][tex]=>\ y(t)\ =\ \int t.e^{-t}dt\ +\ C[/tex]          [tex]=\ (1-t)e^{-t}\ +\ C[/tex]According to question[tex]y(-1)\ =\ (1-(-1))e^1\ +\ C[/tex][tex]=>\ 0\ =\ 2e\ +\ C[/tex][tex]=>\ C\ =\ -2e[/tex]Now, the complete solution of the given differential equation cab be given by [tex]y(t)\ =\ (1-t)e^{-t}\ -\ 2e[/tex]