Exercise · ZW-SEPARABLE-EQUATION-010 · ★★★ Advanced
Solve \(y'=e^{x+y}\), \(y(0)=0\).
Choose one option.
Show answer
- C. \(y=-\ln(2-e^{x})\)
Explanation
Since \(e^{x+y}=e^{x}e^{y}\), the equation is separable: \(\int e^{-y}dy=\int e^{x}dx\) gives \(-e^{-y}=e^{x}+c\). With \(y(0)=0\), \(c=-2\), so \(e^{-y}=2-e^{x}\) and \(y=-\ln(2-e^{x})\).
Review the concept: Separable equation
A separable equation is a first-order differential equation whose derivative is the product of a function \(g\) of \(x\) alone and a function \(h\) of \(y\) alone.
\[\frac{dy}{dx}=g(x)\,h(y)\;\Longrightarrow\;\int\frac{dy}{h(y)}=\int g(x)\,dx\]
More practice problems: separable equation
- ZW-SEPARABLE-EQUATION-001 Which of these equations is separable?
- ZW-SEPARABLE-EQUATION-002 When separating variables in y'=2xy, which equality of integrals results?
- ZW-SEPARABLE-EQUATION-003 Find the general solution of y'=3y.
- ZW-SEPARABLE-EQUATION-004 Separating variables in y'=x/y, which implicit relation do the solutions satisfy?
- ZW-SEPARABLE-EQUATION-005 Solve the problem y'=y, y(0)=2.
- ZW-SEPARABLE-EQUATION-006 Solve y'=-2xy, y(0)=3.
- ZW-SEPARABLE-EQUATION-007 Solve y'=y^2, y(0)=1.
- ZW-SEPARABLE-EQUATION-008 When solving y'=y(y-2) we divide by y(y-2). Which constant solutions might be lost?