Reduction of Order — Question 3

Question 3 – Solution

Strategy. Keep the definite integral and use the fundamental theorem of calculus for verification and bounds.

Step 1: Reduce. The seed has derivatives $2te^{t^2}$ and $(2+4t^2)e^{t^2}$, so its residual vanishes. With $y=e^{t^2}v$, $$w'+2tw=0,\qquad w=v'=Ce^{-t^2}.$$

Step 2: Fit and verify. The zero initial value sets the constant in $v$ to zero; the initial slope sets $C=1$. Therefore $$\boxed{y(t)=e^{t^2}\int_0^t e^{-s^2}\,ds.}$$ Differentiation gives $y'=2ty+1$, then $y''=2y+2ty'$. The residual is zero, and at zero the value and slope are $0,1$.

Step 3: Use symmetry and sign. The integrand is positive and even, so its integral from zero is odd and has the sign of $t$. Multiplication by $e^{t^2}$ preserves both properties. Consequently $ty\ge 0$ and $y'=2ty+1\ge 1$ everywhere: $y$ is strictly increasing.

Step 4: Bound and evaluate. For $0\le s\le t$, $$e^{-t^2}\le e^{-s^2}\le 1\quad\Longrightarrow\quad te^{-t^2}\le\int_0^t e^{-s^2}\,ds\le t.$$ Multiplying by $e^{t^2}$ proves the bounds. Numerical integration gives $y(1)\approx 2.030078$. The integrand is smooth, so the definite integral defines a smooth function on all of $\mathbb R$; the exact residual check proves it solves the equation.