Given that three solutions to the inhomongenous DE equation y''+p(t)y'+q(t)y=g(t)

are y1=1+e^(t^2), y2= 1+te^(t^2), y3= 1+(t+1)e^(t^2),

find the general solution

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- May 21st 2011, 04:28 AMDukeFinding general solution
Given that three solutions to the inhomongenous DE equation y''+p(t)y'+q(t)y=g(t)

are y1=1+e^(t^2), y2= 1+te^(t^2), y3= 1+(t+1)e^(t^2),

find the general solution - May 21st 2011, 10:33 AMJester
Since $\displaystyle y_1$, $\displaystyle y_2$ and $\displaystyle y_3$ all satisfy the ODE then

$\displaystyle u_1 = y_3 - y_2 = e^{t^2}$ and $\displaystyle u_2 = y_3 - y_2 = t e^{t^2}$

satisfy

$\displaystyle u'' + pu' + q u = 0$

Thus, the general solution is

$\displaystyle y = c_1 e^{t^2} + c_2 t e^{t^2} + 1$.

Note that we could have used either $\displaystyle y_1, y_2\; \text{or}\; y_3$ for $\displaystyle y_p$ but can adjust the constants such that all we need is $\displaystyle y_p =1$. - May 21st 2011, 01:06 PMDuke
Why did you subtract the solutions? Thanks

- May 21st 2011, 06:12 PMJester
It was to get to the homogeneous ODE

If $\displaystyle y_1$ and $\displaystyle y_2$ satisfy your ODE, i.e.

$\displaystyle y_1'' + py_1' + q y_1 = g$

and

$\displaystyle y_2'' + py_2' + q y_2 = g$

then substracting gives

$\displaystyle (y_2''-y_1'') + p(y_2'-y_1') + q(y_2-y_1) = 0$.