Question

Find constants A and B such that the function \(y = A\sin x + B\cos x\)satisfies the differential equation\(y'' + y' - 2y = \sin x\).

Short answer

The solution of a differential equation is\(A = \frac{{ - 3}}{{10}},B = \frac{{ - 1}}{{10}}\).

Step-by-step solution

Precise Definition of differential equation

Differential equation used in math it relates to one or more function and their derivative.The differential equation consists of at least one derivative either in the form of an ordinary derivative or partial derivative.

Simplify the obtained condition

To find the constant A and B the function \(y = A\sin x + B\cos x\)which satisfies the differential equation \(y'' + y' - 2y = \sin x\).

The first and second derivativesof \(y\) are:

\(\begin{array}{l}y' = A\cos x - B\sin x\\y'' = - A\sin x - B\cos x\end{array}\)

Put the value of \(y\),\(y'\) and \(y''\) in the differential equation.

\(\begin{array}{c} - A\sin x - B\cos x + A\cos x - B\sin x - 2A\sin x - 2B\cos x = \sin x\\( - 3A - B)\sin x + (A - 3B)\cos x = \sin x\end{array}\)

Compare both sides,

\(\begin{array}{c} - 3A - B = 1\;\;\;\;\; \ldots \left( 1 \right)\\A - 3B = 0\;\;\;\; \ldots \left( 2 \right)\end{array}\)

Find constant

Multiply equation (2) by 3 then add the obtained equation to equation (1).

\(\begin{aligned} - 3A - B + 3A - 9B &= 1 + 0\\ - 10B &= 1\\B &= - \frac{1}{{10}}\end{aligned}\)

Multiply equation (1) by 3 then Subtractequation (2) from the obtained equation.

\(\begin{aligned} - 9A - 3B - A + 3B &= 3 + 0\\ - 10A &= 3\\A &= - \frac{1}{{10}}\end{aligned}\)

So, \(A = \frac{{ - 3}}{{10}},B = \frac{{ - 1}}{{10}}\)is the final answer of this differential equation.