how to find solution of 2nd order differential equations without trial and error? With only the skills learn in Cal I

Question

How do we solve a question like this with only the skills in Calculus 1. Or is trial and error our best shot?

I tried separating it to one variable on each side, but ended up with $y \ln y + y + c = \frac {x^2}{2} +c$ and couldn't reduce further to solve for y...

Just take the second derivative of each one and see what you get...? — Hans Lundmark, Aug 22 '21 at 05:48
It depends on exactly what you mean by "Calculus I", which I don't know. Of course if you know how solve constant-coefficient linear ODEs using the characteristic polynomial, you could do that and see which option doesn't match the form of the general solution. — Hans Lundmark, Aug 22 '21 at 05:51
I don't think I've learnt ODEs. Could I continue from what I tried? I separated them one variable to each side and then integrated it back to this: $y \ln y + y + c = \frac {x^2}{2} +c$ but can't seem to go on from there — nvs0000, Aug 22 '21 at 05:54
That's not at all correct! What on earth did you do there? Something similar to this, or what? — Hans Lundmark, Aug 22 '21 at 05:59
@HansLundmark, from khan academy, they said at this stage we can treat them as fractions.. — nvs0000, Aug 22 '21 at 09:30
Where? I'm pretty sure they did not say that for an ODE like this! For first-order separable equations $dy/dx=F(x) , G(y)$ you can sort of do that, although that explanation is a source of endless confusion and hides that fact that what you are really doing is just using the chain rule backwards. But for second-order ODEs it doesn't work like that – why would it? — Hans Lundmark, Aug 22 '21 at 09:57
Have a look at this unit on Khan Academy instead; the very first item is enough for your case $y''-3y=0$. — Hans Lundmark, Aug 22 '21 at 10:21

score 1 · Answer 1 · answered Aug 22 '21 at 13:08

You ask how to solve this problem using only Calculus I skills, not how to solve this equation. The problem does not ask you to solve the equation, it offers several functions and asks which, if any, is not a solution to the equation. That can be done by differentiating each function and plugging that into the equation. That is NOT "trial and error"!

how to find solution of 2nd order differential equations without trial and error? With only the skills learn in Cal I

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