Linear Equations — Question 3

Question 3 – Solution

Strategy. The integrating factor must match the signed coefficient of $y$. Check its product derivative before integrating.

Step 1: Diagnose the sign error. The student’s derivative is $$(e^{2x}y)'=e^{2x}(y'+2y),$$ whereas multiplying the original equation by $e^{2x}$ gives $e^{2x}(y'-2y)$. These expressions differ in the sign of the $y$ term. The required coefficient is $p=-2$, so $\boxed{\mu=e^{-2x}}$.

Step 2: Correct integration. Multiplication gives $$(e^{-2x}y)'=x,\qquad e^{-2x}y=\frac{x^2}{2}+C.$$ Since $y(0)=0$, $C=0$, and $$\boxed{y=\frac{x^2}{2}e^{2x},\qquad x\in\mathbb R.}$$

Step 3: Verify. Differentiating explicitly, $$y'=xe^{2x}+x^2e^{2x},\qquad y'-2y=xe^{2x}+x^2e^{2x}-x^2e^{2x}=xe^{2x}.$$ The initial value is also zero.

Step 4: Vary the initial datum. The general initial value sets $C=a$, giving $$\boxed{y_a(x)=\left(a+\frac{x^2}{2}\right)e^{2x}.}$$ For distinct $a,b$, $$y_a(x)-y_b(x)=(a-b)e^{2x}\ne 0$$ at every real $x$. Hence their graphs never intersect. The nonintersection follows from the explicit homogeneous difference, not merely from a sketch.