Linear Equations — Question 6

Question 6

For each real initial value $a$, consider $$y'+2xy=1,\qquad y(0)=a.$$ A definite integral is an acceptable exact expression; no elementary antiderivative is required.

Tasks

1. Use an integrating factor to obtain an exact solution based at $x=0$, and state its interval of definition.

2. Verify the formula by differentiation and the initial condition directly.

3. For solutions with initial values $a$ and $b$, determine their difference, its greatest absolute value on $\mathbb R$, and whether their curves can cross when $a\ne b$.

4. Show that the solution with $a=0$ is odd and positive for $x>0$. Sketch the curves for $a=-1,0,1$ in your solution.

Question 6 – Solution

Strategy. Use a definite integral to incorporate the initial value exactly. The homogeneous difference makes comparison easier than evaluating the integral.

Step 1: Integrating factor and solution. Here $$\mu=e^{\int 2x\,dx}=e^{x^2},\qquad (e^{x^2}y)'=e^{x^2}.$$ Integrating from $0$ to $x$ gives $$\boxed{y_a(x)=e^{-x^2}\left(a+\int_0^x e^{t^2}\,dt\right),\qquad x\in\mathbb R.}$$ The integral exists for every finite real $x$.

Step 2: Direct verification. The product rule and Fundamental Theorem give $y_a'=-2xy_a+e^{-x^2}e^{x^2}=1-2xy_a$. Also the integral is zero at $0$, so $y_a(0)=a$.

Step 3: Compare initial values. The integral cancels in the difference: $$\boxed{y_a-y_b=(a-b)e^{-x^2},\qquad \max_{x\in\mathbb R}|y_a-y_b|=|a-b|.}$$ For $a\ne b$, equality in the maximum occurs only at $x=0$, and the difference never changes sign or vanishes at a finite point. The curves cannot cross.

Step 4: Symmetry of the zero-data solution. Since $e^{t^2}$ is even, $\int_0^x e^{t^2}dt$ is odd. Multiplying by the even factor $e^{-x^2}$ preserves oddness. For $x>0$, both the integral and prefactor are positive, so $y_0(x)>0$.