Question

Solve the equation (D^4-4)Y=X²cos h2x​

Answer 1

Step-by-step explanation:

find special solution of (D2+4)y=sinx

yc=c1sin2x+c2cos2x

yp=v1sin2x+v2cos2x

v′1sin2x+v′2cos2x=0(1)

2v′1cos2x−2v′2sin2x=sinx(2)

I tried to solve these two equation and found v1=−cos3x12+cosx4 couldt continue and also answer is given y=−xcos2x4. it seems my way is wrong or the answer.

and from inverse operator y might be

y=1F(−12)sinx=13sinx

Answer 2

Answer:

$\tan( \beta e\pi\pi \beta \gamma \gamma 22.2 \times 32 - .33 = \div 5.266323 {25x555}^{2} \times \frac{?}{?} )$

Step-by-step explanation:

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