Question

Solve \(e^{3x}\,dx+e^{4y}\,dy=0\).

• A. \(e^{3x+4y}=k\)
• B. \(e^{3x}+e^{4y}=k\)
• C. \(\dfrac{1}{3}e^{3x}+\dfrac{1}{4}e^{4y}=k\)
• D. \(e^{3x}+e^{4y}+e^{3x+4y}=k\)

Hint:

\(\int e^{ax}\,dx=\dfrac{1}{a}e^{ax}\). Add constants into a single \(k\).

Answer 1

The Correct Option is C

Idea. Again the equation is separable: one pure \(x\) term and one pure \(y\) term. Move one to the other side and integrate.
\[ e^{3x}dx=-e^{4y}dy $\Rightarrow$ \int e^{3x}dx+\int e^{4y}dy=C $\Rightarrow$ \frac{1}{3}e^{3x}+\frac{1}{4}e^{4y}=k. \]