Substitutions — Question 7

Question 7

Consider $$y'=xe^{-y}-1,\qquad y(0)=0.$$

Tasks

1. Set $u=e^y$ and derive a linear differential equation for $u$.

2. Solve it using the initial condition and recover $y$.

3. Prove that the selected transformed solution stays positive for every real $x$. Find the minimum value of $y$ and where it occurs.

4. The same linear equation has the solution $u=x-1$. Determine precisely where it produces a real original solution, and explain why a valid transformed solution need not be invertible everywhere.

Question 7 – Solution

Strategy. Exponentiating the unknown makes the reciprocal exponential linear, but the inverse logarithm requires a positive transformed value.

Step 1: Transform and solve. For $u=e^y$, $u'=e^y y'=x-u$. The initial value is $u(0)=1$, so $$u'+u=x,\qquad (e^x u)'=xe^x,$$ which gives $u=x-1+Ce^{-x}$. The data select $C=2$, hence $$\boxed{y(x)=\ln(x-1+2e^{-x})}.$$

Step 2: Establish the inverse domain and minimum. Let $U(x)=x-1+2e^{-x}$. Then $U'=1-2e^{-x}$ and $U''=2e^{-x}>0$. Its unique global minimum occurs at $x=\ln 2$ and equals $\ln 2>0$. Therefore the logarithm is defined and smooth for all real $x$, so the IVP interval is $\boxed{\mathbb R}$.

Since the logarithm is increasing, the solution has unique minimum $$\boxed{y(\ln 2)=\ln(\ln 2)}.$$ Verification is immediate from $U'=x-U$: $y'=U'/U=x/U-1=xe^{-y}-1$, and $y(0)=\ln 1=0$.

Step 3: Test another transformed solution. The linear solution $u=x-1$ is defined everywhere, but $u>0$ only for $x>1$. It gives $$\boxed{y=\ln(x-1),\qquad x>1}.$$ On that interval, $y'=1/(x-1)=x/(x-1)-1$, verifying the original equation. At $x\downarrow 1$, $y\to-\infty$, and for $x\le 1$ no real logarithm of $u$ exists. The map $y\mapsto e^y$ has range $(0,\infty)$; solving the transformed linear equation does not remove that range restriction.