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EXERCISE 1.5 · Q194

Q.y=eaxsin⁡(bx)y=e^{ax}\sin(bx), show that y2−2ay1+(a2+b2)y=0y_2-2ay_1+(a^2+b^2)y=0 (where y1=dy/dxy_1=dy/dx, y2=d2y/dx2y_2=d^2y/dx^2)

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Given y=eaxsin⁡(bx)y=e^{ax}\sin(bx).

Step 1: y1=aeaxsin⁡bx+beaxcos⁡bxy_1=ae^{ax}\sin bx+be^{ax}\cos bx.

Step 2: y2=a(aeaxsin⁡bx+beaxcos⁡bx)+(abeaxcos⁡bx−b2eaxsin⁡bx)y_2=a\left(ae^{ax}\sin bx+be^{ax}\cos bx\right)+\left(abe^{ax}\cos bx-b^2e^{ax}\sin bx\right)

=eax[(a2−b2)sin⁡bx+2abcos⁡bx]=e^{ax}\left[(a^2-b^2)\sin bx+2ab\cos bx\right]

Step 3 — compute y2−2ay1y_2-2ay_1: y2−2ay1=eax[(a2−b2)sin⁡bx+2abcos⁡bx]−2a eax[asin⁡bx+bcos⁡bx]y_2-2ay_1=e^{ax}[(a^2-b^2)\sin bx+2ab\cos bx]-2a\,e^{ax}[a\sin bx+b\cos bx] …

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