Linear Equations — Question 5

Question 5 – Solution

Strategy. Subtract the two equations to eliminate the forcing. A nonzero solution difference reveals the coefficient $p$.

Step 1: Recover the coefficients. Let $d=y_1-y_2=2e^{-x}$. Subtraction gives $d'+pd=0$, so $$p=-\frac{d'}d=1,$$ valid for every real $x$ because $d$ never vanishes. Using $y_1'=1-e^{-x}$, $$q=y_1'+y_1=1-e^{-x}+x+e^{-x}=x+1.$$ Thus $\boxed{y'+y=x+1}$ is uniquely recovered within the stated standard linear form.

Step 2: All solutions. The integrating factor is $e^x$, giving $$(e^xy)'=(x+1)e^x=(xe^x)'.$$ Integration yields $\boxed{y=x+Ce^{-x},\quad C\in\mathbb R}$. Substitution confirms $y'+y=x+1$ for every constant.

Step 3: Which combinations work? Write $L[v]=v'+v$. Linearity gives $$L[\alpha y_1+\beta y_2]=(\alpha+\beta)(x+1).$$ For this to equal $x+1$ throughout an interval, we require $$\boxed{\alpha+\beta=1.}$$ These combinations are $x+(\alpha-\beta)e^{-x}$ and generate every solution.

Step 4: Nonhomogeneous versus homogeneous. The sum $y_1+y_2=2x$ produces $L[2x]=2x+2$, not $x+1$. The difference instead satisfies $L[y_1-y_2]=q-q=0$. Thus differences solve the homogeneous equation; combinations with coefficients summing to one preserve the original forcing.