# Find the solution of the following Differential Equations (2x+e^{y})dx + xe^{y}

Find the solution of the following Differential Equations
$$\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

nghodlokl
Step 1
$$\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0}$$
$$\displaystyle\Rightarrow{2}{x}{\left.{d}{x}\right.}+{e}^{{{y}}}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0}$$
$$\displaystyle\Rightarrow{2}{x}{\left.{d}{x}\right.}+{d}{\left({x}{e}^{{{y}}}\right)}={0}$$
$$\displaystyle\therefore{d}{\left({f}\cdot{g}\right)}={f}{d}{\left({g}\right)}+{g}{d}{\left({f}\right)}$$
Step 2
$$\displaystyle\Rightarrow\int{2}{x}{\left.{d}{x}\right.}+\int{d}{\left({x}{e}^{{{y}}}\right)}={0}$$
$$\displaystyle\Rightarrow{2}{\frac{{{x}^{{{2}}}}}{{{2}}}}+{x}{e}^{{{y}}}={c}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}+{x}{e}^{{{y}}}={c}$$
###### Not exactly what youâ€™re looking for?
Solution:
$$\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0},{y}{\left({1}\right)}={0}$$
Compare above with $$\displaystyle{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}={0}$$
$$\displaystyle{M}={2}{x}+{e}^{{{y}}},{N}={x}{e}^{{{y}}}$$
$$\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={e}^{{{y}}},{\frac{{\partial{N}}}{{\partial{x}}}}={e}^{{{y}}}$$
Since $$\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{x}}}}$$, therefore differential equation is exact
$$\displaystyle{\frac{{\partial{f}}}{{\partial{x}}}}={2}{x}+{e}^{{{y}}}$$ (i)
$$\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={x}{e}^{{{y}}}$$ (ii)
integrate (i) wrt x, we get
$$\displaystyle{f{{\left({x},{y}\right)}}}=\int{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{2}}}+{x}{e}^{{{y}}}+{h}{\left({y}\right)}$$
Differentiate above wrt y, we get
$$\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={x}{e}^{{{y}}}+{h}'{\left({y}\right)}$$
compare above and (ii), we get
$$\displaystyle{x}{e}^{{{y}}}+{h}'{\left({y}\right)}={x}{e}^{{{y}}}$$
$$\displaystyle{h}'{\left({y}\right)}={0}$$
integrate both sides wrt y, we get $$\displaystyle{h}{\left({y}\right)}={C}$$
therefore, we get
$$\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{2}}}+{x}{e}^{{{y}}}+{C}$$
or we can write as $$\displaystyle{x}^{{{2}}}+{x}{e}^{{{y}}}+{C}={0}$$
karton

Consider the differential equation
$$(2x+e^{y})dx+xe^{y}dy=0$$
Compare it with $$M(x,y)dx+N(x,y)dy=0$$, get
$$M(x,y)=2x+e^{y}\ \text{and}\ N(x,y)=xe^{y}$$
$$\frac{\partial M}{\partial y}=e^{y}, \frac{\partial N}{\partial x}=e^{y}$$
Observe that, $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
So, the given differential equation is in exact form
$$(2x+e^{y})dx+xe^{y}dy=0$$
$$2xdx+e^{y}dx+xe^{y}dy=0$$
$$2xdx+d(xe^{y})=0$$
on taking integration get
$$x^{2}+xe^{y}=C$$
which is the required solution.