# Solve dy/dx=xye^(3x) ?

Mar 20, 2017

y=1/9e^(1/9e^(3x)(3x-1)

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x y {e}^{3 x}$

$\int \frac{1}{y} \mathrm{dy} = \int x {e}^{3 x} \mathrm{dx}$

$\ln | y | = \frac{1}{9} {e}^{3 x} \left(3 x - 1\right) + \ln k$

$\ln 1 = \frac{1}{9} {e}^{0} \left(3 \left(0\right) - 1\right) + \ln k$

$k = \frac{1}{9}$

$\ln | y | - \ln \left(\frac{1}{9}\right) = \frac{1}{9} {e}^{3 x} \left(3 x - 1\right)$

$\ln | 9 y | = \frac{1}{9} {e}^{3 x} \left(3 x - 1\right)$

9y=e^(1/9e^(3x)(3x-1)

y=1/9e^(1/9e^(3x)(3x-1)

Mar 20, 2017

$y = {e}^{\left(\frac{x}{3} - \frac{1}{9}\right) {e}^{3 x} + \frac{1}{9}}$

#### Explanation:

This is a separable differential equation. It can be arranged as

${f}_{1} \left(x\right) \mathrm{dx} = {f}_{2} \left(y\right) \mathrm{dy}$

with ${f}_{1} \left(x\right) = x {e}^{3 x}$ and ${f}_{2} \left(y\right) = \frac{1}{y}$

Integrating we have

${F}_{1} \left(x\right) + C = {F}_{2} \left(y\right)$

with

${F}_{1} \left(x\right) = \left(\frac{x}{3} - \frac{1}{9}\right) {e}^{3 x}$ and

${F}_{2} \left(y\right) = \log \left(y\right)$

so

$\log \left(y\right) = \left(\frac{x}{3} - \frac{1}{9}\right) {e}^{3 x} + C$ or

$y = {C}_{1} {e}^{\left(\frac{x}{3} - \frac{1}{9}\right) {e}^{3 x}}$

solving for initial conditions

$1 = {C}_{1} {e}^{\left(\frac{0}{3} - \frac{1}{9}\right) {e}^{3 \times 0}}$ giving

${C}_{1} = {e}^{\frac{1}{9}}$ and finally

$y = {e}^{\frac{1}{9}} {e}^{\left(\frac{x}{3} - \frac{1}{9}\right) {e}^{3 x}} = {e}^{\left(\frac{x}{3} - \frac{1}{9}\right) {e}^{3 x} + \frac{1}{9}}$