Question

(D²-D²-D-1) y = 4sinx
please help me ​

Answer 1

Answer:

First, this is a third-order ODE, not first-order as stated in the question.

The general solution to this differential equation is the sum of the homogeneous solution, yh , and the particular solution, yp .

To find the homogeneous solution, set the characteristic polynomial equal to zero and solve for its roots. The characteristic polynomial is:

P(s)=s3−s2+s−1

P(s)=(s−1)(s2+1)

Setting this equal to zero and solving for the roots, we find:

s=1

s=±i

Therefore, the homogeneous solution is:

yh=c1ex+c2sin(x)+c3cos(x)

To find the particular solution, we will use the exponential response formula. Since P(i)=0 , the particular solution is the imaginary part of:

zp=4xeixP′(i)

where P′(i)=3(i)2−2(i)+1=−2(1+i)

zp=4xeix−2(1+i)

zp=−xeix(1−i)

zp=−x(cos(x)+isin(x))(1−i)

yp=Im(zp)=x(cos(x)−sin(x))

Therefore, the general solution is:

y=yh+yp

y=c1ex+c2sin(x)+c3cos(x)+x(cos(x)−sin(x))

Answer 2

Answer:

ohh thanks for .

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