Questions: Find the general solution of the differential equation. (Enter your solution as an equation.) y' e^(y-6x)=e^(x+2y)

Solution Steps

We start with the given differential equation:

\[ y' e^{y - 6x} = e^{x + 2y} \]

Rearranging this, we can express \(y'\) as:

\[ y' = e^{x + 2y} e^{-(y - 6x)} = e^{7x + y} \]

Next, we separate the variables \(y\) and \(x\):

\[ \frac{dy}{dx} = e^{7x + y} \]

This can be rewritten as:

\[ \frac{dy}{e^y} = e^{7x} dx \]

We integrate both sides:

\[ \int e^{-y} dy = \int e^{7x} dx \]

The left side integrates to:

\[ -e^{-y} \]

And the right side integrates to:

\[ \frac{1}{7} e^{7x} + C \]

Setting the two integrals equal gives us:

\[ -e^{-y} = \frac{1}{7} e^{7x} + C \]

Multiplying through by \(-1\) yields:

\[ e^{-y} = -\frac{1}{7} e^{7x} - C \]

Taking the natural logarithm of both sides results in:

\[ -y = \ln\left(-\frac{1}{7} e^{7x} - C\right) \]

Finally, we solve for \(y\):

\[ y = -\ln\left(-\frac{1}{7} e^{7x} - C\right) \]

This expression represents the general solution of the differential equation.

Final Answer