Question

Solve the following differential equation d^2y/dx^2 - 2dy/dx + 10y = 0,. Given Y(0) = 4. dy/dx (0) = 1

Answer 1

$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+10y=0\\ \text{The auxilliary equation is going to be}\\ m^2-2m+10=0\\ \text{Using quadratic formula}\\ m=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(10)}}{2(10)}\\ m=\frac{2\pm\sqrt{-36}}{2}\\ \text{m1 = 1+3i and m2 = 1- 3i}\\ Y=A_1e^{(1+3i)x}+A_2e^{(1-3i)x}\\ Y=A_1e^x.e^{3ix}+A_2e^x.e^{-3ix}\\ Y=A_1e^x[cos(3x)+isin(3x)]+A_2e^x[cos(3x)-isin(3x)]\\ Y=[(A_1+A_2)e^xcos(3x)]+[(A_1-A_2)e^xisin(3x)]\\ Let (A_1+A_2)=A\\ (A_1-A_2)=B \implies\\ Y=e^x[Acos(3x)+Bisin(3x)]\\ \text{since Y(0)=4, then}\\ Y(0)=e^0[Acos3(0)+Bisin3(0)]\\ \text{4=A that is}\\ \text{A = 4}\\ \frac{dy}{dx}=Ae^x[cos(3x)-3isin(3x)]+Be^x[3cos(3x)+isin(3x)]\\ \frac{dy}{dx}(0)=Ae^0[cos(3(0))-3isin(3(0))]+Be^0[3cos(3(0))+isin(3(0))]\\ \frac{dy}{dx}(0)=A+3B\\ 1= 4 +3B\\ 3B=-3\\ B=-1\\ Y=e^x[4cos(3x)-isin(3x)]$