$$ f''(x)+g'(k)f'(x)+g(k)f(x)=A $$
k is a constant, solve for f(x)

Question

$$ f''(x)+g'(k)f'(x)+g(k)f(x)=A $$
k is a constant, solve for f(x)

experimentx · Answer

*
BTW where did you get it?? do you have answer for it?

anonymous · Answer

No.

$$(D^2+aD+b)f(x)e^{kx}=Ae^{kx}$$

anonymous · Answer

Give$$f''(x)+f'(x)(2k+a)+f(x)(k^2+ak+b)=A$$s

experimentx · Answer

these two are different ... you have to use product rule here.

anonymous · Answer

That was after simplifying- it didn't jump straight to that

experimentx · Answer

how did you put g(x) = e^kx??

anonymous · Answer

g(x)=k^2+ak+b

experimentx · Answer

so your y was e^kx f(x) given at beginning?

anonymous · Answer

(f'' + kf' + f'k + k^2 f)e^{kx}+(af' + akf)e^{kx} + bfe^{kx}= A e^{kx}
(f'' + f'(2k+a) +f(k^2 + ak +b) )e^{kx} = A e^{kx}

experimentx · Answer

was this 
$$ (D^2+aD+b)f(x)e^{kx}=Ae^{kx} $$
given at beginning?

anonymous · Answer

Yes

anonymous · Answer

Sorry for being a little cryptic

experimentx · Answer

$$ f''(x)+f'(x)(2k+a)+f(x)(k^2+ak+b)=A $$
this looks like exact differential equation ... let me check my book.

anonymous · Answer

With the 2 variables x and k? Clever, but I've never needed encountered one in the wild before, so I've forgotten the little I've learnt.

experimentx · Answer

here is definition of exact differential equation

anonymous · Answer

On what basis is it exact?

experimentx · Answer

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