Step-by-step Solution

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

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Final Answer

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

Problem to solve:

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

Choose the solving method

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 of the equality

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

$-\frac{1}{2}\left(\frac{1}{e^u}\right)$

Multiply the fraction and term

$\frac{-\frac{1}{2}}{e^u}$

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

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

Final Answer

$y=\frac{\ln\left(\frac{-\frac{1}{2}}{\frac{1}{3}e^{3x}+C_0}\right)}{2}$
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8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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