Answer to Question #301902 in Differential Equations for haru

Question

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

Answer 1

Let us write the characteristic equation for this DE:

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

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

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

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

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