Question #301902

solve the initial value problems y''-2y'+2y=0,y(0)=0,y'(0)=1


1
Expert's answer
2022-02-24T15:03:53-0500

Let us write the characteristic equation for this DE:

λ22λ+2=0,λ1=1i,  λ2=1+i.\lambda^2 - 2\lambda + 2 = 0, \\ \lambda_1 = 1-i, \; \lambda_2 = 1+i.

So the solution of the equation will be y=y1+y2=C1exsin(x)+C2excos(x).y = y_1 + y_2 = C_1 e^{x}\sin(x) + C_2e^x\cos(x).

y(0)=0        C11sin(0)+C21cos(0)=0+C2=0,C2=0.y(0) = 0 \;\; \Rightarrow \;\; C_1 \cdot1\cdot\sin(0) + C_2 \cdot1\cdot\cos(0) =0 + C_2 = 0, \\ \Rightarrow C_2 = 0.

y(0)=1        (C1exsin(x))=C1exsin(x)+C1excos(x),(C1exsin(x))x=0=C1=1.y'(0) = 1 \;\; \Rightarrow \;\; (C_1e^x\sin(x))' = C_1e^x\sin(x) + C_1e^x\cos(x), \\ (C_1e^x\sin(x))'|_{x=0} = C_1 = 1.

The final solution is y=exsin(x).y=e^x\sin(x).


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