UPSC 2017 Maths Optional Paper 1 Q6b-ii — Step-by-Step Solution

7 marks · Section B

Linear ODE with constant coefficients · ODEs · asked 4× in 14 yrs · Read the full method →

Question

Solve the following initial value differential equations: $20y''+4y'+y=0,\ y(0)=3.2$ and $y'(0)=0$ .

Constant-coefficient homogeneous ODE; complex roots $\Rightarrow$ $e^{\alpha x}(A\cos\beta x+B\sin\beta x)$ ; apply the two ICs.

20m^2+4m+1=0\ \Rightarrow\ m=\frac{-4\pm\sqrt{16-80}}{40}=\frac{-4\pm\sqrt{-64}}{40}=\frac{-4\pm 8i}{40}=-\frac1{10}\pm\frac{i}{5}.

Complex roots $\alpha\pm i\beta$ with $\alpha=-\tfrac1{10},\ \beta=\tfrac15$ .

y=e^{-x/10}\Big(A\cos\tfrac{x}{5}+B\sin\tfrac{x}{5}\Big).

At $x=0$ : $y(0)=A=3.2$ .

Differentiate:

y'=e^{-x/10}\Big[-\tfrac1{10}\big(A\cos\tfrac{x}5+B\sin\tfrac{x}5\big)+\big(-\tfrac{A}{5}\sin\tfrac{x}5+\tfrac{B}{5}\cos\tfrac{x}5\big)\Big].

At $x=0$ : $y'(0)=-\tfrac{A}{10}+\tfrac{B}{5}=0\ \Rightarrow\ B=\tfrac{A}{2}=1.6.$

\boxed{\;y=e^{-x/10}\Big(3.2\cos\tfrac{x}{5}+1.6\sin\tfrac{x}{5}\Big)=\frac{8}{5}\,e^{-x/10}\Big(2\cos\tfrac{x}{5}+\sin\tfrac{x}{5}\Big).\;}