π¦(π₯) = π1ππ₯ + π2πβπ₯ + 4π ππ(π₯)
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"
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