# Step-by-step Solution

## Solve the differential equation $\frac{dy}{dx}=e^{\left(3x+2y\right)}$

Go!
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$y=\frac{\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right)}{2}$

## Step-by-step Solution

Problem to solve:

$\frac{dy}{dx}=e^{3x+2y}$
1

Apply the property of the product of two powers of the same base in reverse: $a^{m+n}=a^m\cdot a^n$

$\frac{dy}{dx}=e^{3x}e^{2y}$
2

Group the terms of the differential equation. Move the terms of the $y$ variable to the left side, and the terms of the $x$ variable to the right side

$\frac{1}{e^{2y}}dy=e^{3x}dx$
3

Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$

$\int\frac{1}{e^{2y}}dy=\int e^{3x}dx$

We can solve the integral $\int\frac{1}{e^{2y}}dy$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $2y$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=2y$

Now, in order to rewrite $dy$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2dy$

Isolate $dy$ in the previous equation

$\frac{du}{2}=dy$

Substituting $u$ and $dy$ in the integral and simplify

$\frac{1}{2}\int\frac{1}{e^u}du$

Apply the formula: $\int\frac{1}{e^x}dx$$=-\left(\frac{1}{e^x}\right), where x=u \frac{-\frac{1}{2}}{e^u} Replace u with the value that we assigned to it in the beginning: 2y \frac{-\frac{1}{2}}{e^{2y}} 4 Solve the integral \int\frac{1}{e^{2y}}dy and replace the result in the differential equation \frac{-\frac{1}{2}}{e^{2y}}=\int e^{3x}dx We can solve the integral \int e^{3x}dx by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it u), which when substituted makes the integral easier. We see that 3x it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part u=3x Now, in order to rewrite dx in terms of du, we need to find the derivative of u. We need to calculate du, we can do that by deriving the equation above du=3dx Isolate dx in the previous equation \frac{du}{3}=dx Substituting u and dx in the integral and simplify \int\frac{e^u}{3}du Take the constant \frac{1}{3} out of the integral \frac{1}{3}\int e^udu The integral of the exponential function is given by the following formula \displaystyle \int a^xdx=\frac{a^x}{\ln(a)}, where a > 0 and a \neq 1 \frac{1}{3}e^u Replace u with the value that we assigned to it in the beginning: 3x \frac{1}{3}e^{3x} 5 Solve the integral \int e^{3x}dx and replace the result in the differential equation \frac{-\frac{1}{2}}{e^{2y}}=\frac{1}{3}e^{3x} 6 As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C \frac{-\frac{1}{2}}{e^{2y}}=\frac{1}{3}e^{3x}+C_0 Take the reciprocal of both sides of the equation \frac{e^{2y}}{-\frac{1}{2}}=\frac{1}{\frac{1}{3}e^{3x}+C_0} Multiply both sides of the equation by -\frac{1}{2} e^{2y}=\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0} We can take out the unknown from the exponent by applying natural logarithm to both sides of the equation \ln\left(e^{2y}\right)=\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right) Apply the formula: \ln\left(e^x\right)$$=x$, where $x=2y$

$2y=\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right)$

Divide both sides of the equation by $2$

$y=\frac{\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right)}{2}$
7

Find the explicit solution to the differential equation

$y=\frac{\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right)}{2}$

$y=\frac{\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right)}{2}$
$\frac{dy}{dx}=e^{3x+2y}$