since we have y=e2x sinxwe gety′=2e2x sin x+e2x cos xy′′=4e2x sin x+2e2x cos x+2e2x cos x−e2x sin x→(1)Also, we have−4y′=−8e2x sin x−4e2x cos x→(2)5y=5e2x sin x→(3)Adding (1), (2), (3), we get y′′−4y′+5y=0\text{since we have }\\ y=e^{2x} \ sin x\\ \text{we get}\\ y^\prime=2e^{2x} \ sin \ x+e^{2x} \ cos \ x\\ y ^{\prime\prime}=4e^{2x} \ sin \ x+2 e^{2x} \ cos \ x\\ +2 e^{2x} \ cos \ x-e^{2x} \ sin \ x\rightarrow(1)\\ \text{Also, we have}\\ -4y^\prime=-8e^{2x} \ sin \ x-4e^{2x} \ cos \ x\rightarrow (2)\\ 5 y=5e^{2x} \ sin \ x\rightarrow (3)\\ \text{Adding (1), (2), (3), we get }\\ y^{\prime \prime}-4y^\prime+5y=0since we have y=e2x sinxwe gety′=2e2x sin x+e2x cos xy′′=4e2x sin x+2e2x cos x+2e2x cos x−e2x sin x→(1)Also, we have−4y′=−8e2x sin x−4e2x cos x→(2)5y=5e2x sin x→(3)Adding (1), (2), (3), we get y′′−4y′+5y=0
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