Answer to Question #261009 in Differential Equations for blossomqt

Question #261009

𝑦(π‘₯) = 𝑐1𝑒π‘₯ + 𝑐2π‘’βˆ’π‘₯ + 4𝑠𝑖𝑛(π‘₯)

1
Expert's answer
2021-11-04T20:13:37-0400

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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