Find constants A and B such that the function y = Asin(x) + Bcos(x) satisfies the differential equation y" + y' - 2y = sin(x)

I can find the derivatives just fine, but I'm not sure where to go in order to solve for A or B.

Find constants A and B such that the function y = Asin(x) + Bcos(x) satisfies the differential equation y" + y' - 2y = sin(x)

I can find the derivatives just fine, but I'm not sure where to go in order to solve for A or B.

I can find the derivatives just fine, but I'm not sure where to go in order to solve for A or B.

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
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Andromeda wrote:Find constants A and B such that the function y = Asin(x) + Bcos(x) satisfies the differential equation y" + y' - 2y = sin(x)

I can find the derivatives just fine, but I'm not sure where to go in order to solve for A or B.

plug your derivatives into the equation and solve for A and B. If you want more help show what you have so far.

y'(x) = Acosx - Bsinx

y"(x) = -Asinx - Bcosx

(-Asinx - Bcosx) + (Acosx - Bsinx) - 2(Asinx + Bcosx) = sinx

From here I'm not sure how to solve for A or B since there are 2 unknowns and only 1 equation

y"(x) = -Asinx - Bcosx

(-Asinx - Bcosx) + (Acosx - Bsinx) - 2(Asinx + Bcosx) = sinx

From here I'm not sure how to solve for A or B since there are 2 unknowns and only 1 equation

- Martingale
**Posts:**335**Joined:**Mon Mar 30, 2009 1:30 pm**Location:**USA-
**Contact:**

Andromeda wrote:y'(x) = Acosx - Bsinx

y"(x) = -Asinx - bcosx

(-Asinx - Bcosx) + (Acosx - Bsinx) - 2(Asinx + Bcosx) = sinx

From here I'm not sure how to solve for A or B since there are 2 unknowns and only 1 equation

Isolate the sines and cosines

you should get something like

or another way to look at it

then

where are functions A and B.

solve this system of two equations and 2 unknowns