For what value of $k$ is $2e^{4x}-5e^{10x}$ a solution to $y''-ky'+40y=0$?

Question

For what value of $k$ is $2e^{4x}-5e^{10x}$ a solution to $y''-ky'+40y=0$?

No clue how to start the question. All I can see is that the characteristic equation for the differential equation shown above looks to be of the form: $r^2 - kr + 40 = 0$.

Answer 1

$$y''-ky'+40y=0$$

$$y=2e^{4x}-5e^{10x}$$ $$y'=8e^{4x}-50e^{10x}$$ $$y''=32e^{4x}-500e^{10x}$$ Substituting the values into the equation, we get $$32e^{4x}-500e^{10x}-8ke^{4x}+50ke^{10x}+80e^{4x}-200e^{10x}=0$$ $$e^{4x}(32-8k+80)-e^{10x}(500-50k+200)=0$$ So comparing the like terms, we get $k=14$.

Answer 2

For such questions, you have to different y to get y' and then differentiate y' to get y'', and arrange the final equation you get in a form similar to that of the question and then compare the coefficients.

$y= 2 e^{4x} - 5e^{10x}$

$y' = 8 e^{4x} - 50 e^{10x}$

$y''= 32 e^{4x} + 500 e^{10x}.$

Hence, substituting these in the equation $y''-ky'+40y$, you can get the value of k.

Answer 3

Taking into account the solution you are given, you already know that the roots of the characteristic equation are $4$ and $10$.

On the other side, the characteristic equation you correctly wrote is $$r^2-k r+40=0$$ and the sum of the roots is $k$ (remember the sum of the roots of a quadratic equation). So ??