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Recall that variation of parameters is used to solve second-order ODEs of the form
y''(t) + p(t) y'(t) + q(t) y(t) = f(t)
so the first thing you need to do is divide both sides of your equation by t :
y'' + (2t - 1)/t y' - 2/t y = 7t
You're looking for a solution of the form
[tex]y=y_1u_1+y_2u_2[/tex]
where
[tex]u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt[/tex]
[tex]u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt[/tex]
and W denotes the Wronskian determinant.
Compute the Wronskian:
[tex]W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}[/tex]
Then
[tex]u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t[/tex]
[tex]u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}[/tex]
The general solution to the ODE is
[tex]y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}[/tex]
which simplifies somewhat to
[tex]\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}[/tex]