Question 3 of 7: Initial Value Problem with Real Distinct Roots and Linear Forcing

Question 3: Initial Value Problem with Real Distinct Roots and Linear Forcing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $y''-2y=-4x$, an unforced-frequency (real root) second-order ODE with linear forcing, $y(0)=2$, $y'(0)=7$.

Find. $y(x)$.

Approach. Solve the homogeneous equation (real roots $\pm\sqrt2$), find a linear particular solution, then apply the initial conditions; express the homogeneous part in $\cosh/\sinh$ form for a compact final answer.

1. Homogeneous solution. $r^2-2=0\Rightarrow r=\pm\sqrt2$: $$y_h=C_1e^{\sqrt2x}+C_2e^{-\sqrt2x}=A\cosh(\sqrt2x)+B\sinh(\sqrt2x).$$
2. Particular solution. Try $y_p=ax+b$ (no constant/linear mode in $y_h$ since roots are $\pm\sqrt2\ne0$). Then $y_p''-2y_p=-2ax-2b=-4x\Rightarrow a=2,\ b=0$: $$y_p=2x.$$
3. Apply initial conditions. $y=A\cosh(\sqrt2x)+B\sinh(\sqrt2x)+2x$, so $y(0)=A=2$. Differentiating, $y'=A\sqrt2\sinh(\sqrt2x)+B\sqrt2\cosh(\sqrt2x)+2$, so $y'(0)=B\sqrt2+2=7\Rightarrow B=\dfrac{5}{\sqrt2}=\dfrac{5\sqrt2}{2}$.

$$y(x)=\boxed{2\cosh(\sqrt2x)+\tfrac{5\sqrt2}{2}\sinh(\sqrt2x)+2x}$$