Linear Equations — Question 7

Question 7 – Solution

Strategy. Integrate the forcing after multiplying by $e^{2x}$, then separate periodic and decaying terms.

Step 1: Integrate. The equation becomes $$(e^{2x}y)'=e^{2x}(3\cos x+\sin x).$$ Seek an antiderivative $e^{2x}(A\cos x+B\sin x)$. Matching coefficients gives $2A+B=3$ and $2B-A=1$, hence $A=B=1$. Indeed, $$[e^{2x}(\cos x+\sin x)]'=e^{2x}(3\cos x+\sin x).$$ Therefore $y=\cos x+\sin x+Ce^{-2x}$. Since $y(0)=0$, $C=-1$, giving $$\boxed{y=\cos x+\sin x-e^{-2x}.}$$

Step 2: The only periodic member. The function $y_p=\cos x+\sin x$ is $2\pi$-periodic and satisfies $y_p'+2y_p=3\cos x+\sin x$. For the general solution, $$y(x+2\pi)-y(x)=Ce^{-2x}(e^{-4\pi}-1).$$ This vanishes for every $x$ only if $C=0$. Hence $\boxed{y_p=\cos x+\sin x}$ is the unique periodic member; the IVP has $C=-1$ and is not periodic.

Step 3: Uniform accuracy after a time. The exact discrepancy is $|y-y_p|=e^{-2x}$, strictly decreasing on $x\ge 0$. Thus $$e^{-2X}\le\frac 1{100}\quad\Longleftrightarrow\quad \boxed{X\ge\ln 10.}$$ The earliest such time is $\ln 10$. The term $Ce^{-2x}$ is transient because it vanishes as $x\to\infty$, leaving the periodic response.