y'-2y=e^(3x), veryfy that the solution is y=e^(3x)+10^(2x).....I cannot find the second term in the solution and have used the integrating factor method but still....please help what wethod should I use...Euler?

Question

anonymous · Answer

I haven't worked it, but try just substituting y and y' into the equation to see if you get an identity at the end.

anonymous · Answer

yeh that's true..lol.rember nooow

anonymous · Answer

I can't get it to verify.

anonymous · Answer

$$y'=3e ^{3x}+(2\ln10)10^{2x}$$
$$[3e ^{3x}+(2\ln10)10^{2x}]-2[e ^{3x}+10^{2x}]=e ^{3x}$$$$e ^{3x}+10^{2x}[(2\ln10)-2]=e ^{3x}$$if 2 ln10 were equal to 2, it would work, but its not

anonymous · Answer

I think that they were counting on you doing the derivative of 10^(2x) wrong.

anonymous · Answer

dnt worry I got it..by substitution!!!

anonymous · Answer

yeah, that's what I did.

anonymous · Answer

I'll leave it at that, but I think it is a trick and that it is not a solution.

anonymous · Answer

yeh...dey full of tricks at ths university i knw..

anonymous · Answer

:)