Question

In Exercises 1–22, solve the differential equation.e⁻ˣ dy + (e⁻ˣ y - 4x) dx = 0

Accepted Answer

Rewrite the given differential equation in the form $$M(x,y)\,dx + N(x,y)\,dy = 0. $$Here, identify $$M(x,y)$$ and $$N(x,y)$$ from the equation $$e^{-x} dy$$ + $$(e^{-x} y - 4x) dx = 0$. Rearranging, we get (e$$^{-x}$$ y - 4x) dx + e$$^{-x}$$ dy = 0$$, so $$M(x,y) = e$$^{-x}$$ y - 4x$$ and $$N(x,y) = e$$^{-x}$$. Check if the differential equation is exact by verifying if $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$. Compute $$\frac{\partial M}{\partial y}$$ and $$\frac{\partial N}{\partial x}$$ separately. If the equation is not exact, find an integrating factor. Since M$ and N$ involve x$ and y$, try to find an integrating factor that depends on x$ or y$ alone by using the formulae for integrating factors for non-exact equations. Once the integrating factor is found, multiply the entire differential equation by it to make the equation exact. Then, solve the exact differential equation by finding a potential function $$\Psi(x,y)$$ such that $$\frac{\partial \Psi}{\partial x} = M$$ and $$\frac{\partial \Psi}{\partial y} = N$$. Integrate M$ with respect to x$ and N$ with respect to y$ to find $$\Psi(x,y)$$, including the function of integration as needed. Finally, set $$\Psi(x,y) = C$$ (a constant) to express the implicit solution to the differential equation.

In Exercises 1–22, solve the differential equation.
e⁻ˣ dy + (e⁻ˣ y - 4x) dx = 0

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