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We have an ode as such:

$y'' + (sint)y'+t^2y=0$

Also, we know that $y_1$ and $y_2$ are linearly independent solutions.

How to show that the general solution has the form: $y=c_1y_1+c_2y_2$

where $c_1,c_2$ are arbitrary constants.

It is not an ode with constant coefficients, in that case it would have $exp(rt)$ as solution. But in this case, to state the requested I would show that $y=c_1y_1+c_2y_2$ also satisfies the ode.

Is this enough to do?

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Simply test $y=c_1y_1+c_2y_2$: \begin{align}y''+\sin(t)y'+t^2y&=(c_1y_1+c_2y_2)''+\sin(t)(c_1y_1+c_2y_2)'+t_2(c_1y_1+c_2y_2)\\&=c_1y_1''+\sin(t)c_1y_1'+t^2c_1y_1+c_2y_2''+\sin(t)c_2y_2'+t^2c_2y_2\\&=c_1(y_1''+\sin(t)y_1'+t^2y_1)+c_2(y_2''+\sin(t)y_2'+t^2y_2)\\&=c_1\cdot 0+c_2\cdot 0\\&=0\end{align} Hence $y=c_1y_1+c_2y_2$ is also a solution of the ODE if $y_1$ and $y_2$ are known solutions.

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Provided that $y_1$ and $y_2$ are solutions:

$$(c_1y_1+c_2y2)''+\sin t (c_1y_1+c_2y2)'+t^2(c_1y_1+c_2y2)$$

$$(c_1y_1''+c_2y2'')+\sin t (c_1y_1'+c_2y2')+t^2(c_1y_1+c_2y2)$$

$$c_1(y_1''+\sin t y_1'+t^2y_1)+c_2(y_2''+\sin t y_2'+t^2y_2)$$

The inside of the parentheses are 0 since $y_1$ and $y_2$ are given solutions

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