A differential equation question

[mech-eng]

Verify that given function is a solution.
y'' - 2y' + 2y = 0 , y=e^x(Acos x + Bsin x)

First I take derivative of y which is y+e^x(-Asin x + B cos x) then I asign e^x(-Asin x + B cos x) to y'-y. Then I take derivative of y' which is y'+(y'-y)-y which equals 2y'-2y=y'' then I use y'' as 2y'-2y which satifies. Do you have a faster or more exciting way?

Have a nice day.

 

[Bill_K]

Just solve the DE. Let y = e^mx. Plug into the DE and get the condition m² -2m + 2 = 0, which has solutions m = 1 ± i.

Thus the DE has solutions y = a e^{(1 + i)x} + b e^{(1 - i)x} = e^x(a e^ix + b e^-ix), which by Euler's formula can be reduced to the form you give.

 

[lurflurf]

That equation begs for the change of variable y=e^x u
y'=e^x(u'+u)
y''=e^x(u''+2u'+u)
so
y'' - 2y' + 2y = 0 , y=e^x(Acos x + Bsin x)
becomes
u''+u=0 , u=Acos x + Bsin x

 

[Bill_K]

The method I gave works for any linear homogeneous DE with constant coefficients, reducing it to an algebraic equation. And in this case all that remains to do is solve a quadratic.

That equation begs for the change of variable y=e^x u

Would you have been able to guess this change of variable if you did not already have the solution to look at?

 

[AlephZero]

The question doesn't ask the OP to solve the equation. It asks to verify the given function is a solution.

Of course this is a very easy ODE to solve if you know how to solve it, but the question might not even be from a differential equations course. It might just be an exercise in differentiation.

 

[lurflurf]

The method I gave works for any linear homogeneous DE with constant coefficients, reducing it to an algebraic equation. And in this case all that remains to do is solve a quadratic.

Not sure what you are getting at. Solving polynomials (other than quadratics) is not easy.

Would you have been able to guess this change of variable if you did not already have the solution to look at?

Sure
m² -2m + 2=(m-1)²+1