Answer to Question #301902 in Differential Equations for haru

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:

"\\lambda^2 - 2\\lambda + 2 = 0, \\\\\n\\lambda_1 = 1-i, \\; \\lambda_2 = 1+i."

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

"y(0) = 0 \\;\\; \\Rightarrow \\;\\; C_1 \\cdot1\\cdot\\sin(0) + C_2 \\cdot1\\cdot\\cos(0) =0 + C_2 = 0, \\\\\n\\Rightarrow C_2 = 0."

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

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


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