Question

Solve the initial value problem

(e^x + y)dx + (2 + x +ye^y)dy=0, y(0)=1

Answer 1

Answer on Question #70217 – Math – Differential Equations

Question

Solve the initial value problem

$$(e^{x} + y)dx + (2 + x + y e^{y})dy = 0, y(0) = 1.$$

Solution

Let $P(x,y) = e^{x} + y$ and $Q(x,y) = 2 + x + y e^{y}$.

This is a total differential equation $P(x,y)dx + Q(x,y)dy = 0$, because $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = 1$.

Define $f(x,y)$ such that $\frac{\partial f}{\partial x} = P(x,y)$ and $\frac{\partial f}{\partial y} = Q(x,y)$. Then the solution will be given by $f(x,y) = c_{1}$, where $c_{1}$ is an arbitrary real constant.

$\frac{\partial f}{\partial x} = e^{x} + y \Rightarrow f(x,y) = \int (e^{x} + y)dx + g(y) = e^{x} + xy + g(y)$ where $g(y)$ is an arbitrary function of $y$.

Differentiate $f(x,y)$ with respect to $y$ in order to find $g(y)$:

$$\begin{array}{l} \frac{\partial f}{\partial y} = x + g'(y) = 2 + x + y e^{y} \Rightarrow g'(y) = 2 + y e^{y} \Rightarrow \\ \Rightarrow g(y) = \int (2 + y e^{y}) dy = 2y + e^{y}(y - 1) + c_{2}. \end{array}$$ $$f(x,y) = e^{x} + xy + 2y + e^{y}(y - 1) + c_{2}.$$ $$e^{x} + xy + 2y + e^{y}(y - 1) + c_{2} = c_{1} \text{ or } e^{x} + xy + 2y + e^{y}(y - 1) = c$$

If $y(0) = 1$ then $e^{0} + 0 * 1 + 2 * 1 + e^{0}(1 - 1) = c \Rightarrow c = 3 \Rightarrow$

$$\Rightarrow e^{x} + xy + 2y + e^{y}(y - 1) = 3$$

Answer: $e^{x} + xy + 2y + e^{y}(y - 1) = 3$.

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