Question

𝑦(𝑥) = 𝑐1𝑒𝑥 + 𝑐2𝑒−𝑥 + 4𝑠𝑖𝑛(𝑥)

Answer 1

Solution;

Eliminate the orbitrary constants as follows:

$y(x)=c_1e^x+c_2e^{-x}+4sin(x)$ ......(a)

Differentiate ;

$y'(x)=c_1e^x-c_2e^{-x}+4cos(x)$ ......(b)

The second derivative is;

$y''(x)=c_1e^x+c_2e^{-x}-4sin(x)$ ....(c)

From equation (c);

$c_1e^x+c_2e^{-x}=y''(x)+4sin(x)$ .....(d)

Substitute (d) into (a):

$y(x)=y''(x)+4sin(x)+4sin(x)$

Hence , equation (a) is an explicit solution of the linear equation;

$y''(x)-y(x)+8sin(x)=0$