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# Solve the differential equation $\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$

## Step-by-step Solution

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$y=-1+\sqrt{2x^2+x^{3}+2x+C_0},\:y=-1-\sqrt{2x^2+x^{3}+2x+C_0}$
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## Step-by-step Solution

Problem to solve:

$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$

Choose the solving method

1

Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$

$2\left(y+1\right)dy-\left(3x^2+4x+2\right)dx=0$
2

The differential equation $2\left(y+1\right)dy-\left(3x^2+4x+2\right)dx=0$ is exact, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and they satisfy the test for exactness: $\displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form $f(x,y)=C$

$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$

Find the derivative of $M(x,y)$ with respect to $y$

$\frac{d}{dy}\left(-\left(3x^2+4x+2\right)\right)$

The derivative of the constant function ($-\left(3x^2+4x+2\right)$) is equal to zero

0

Find the derivative of $N(x,y)$ with respect to $x$

$\frac{d}{dx}\left(2\left(y+1\right)\right)$

The derivative of the constant function ($2\left(y+1\right)$) is equal to zero

0
3

Using the test for exactness, we check that the differential equation is exact

$0=0$

The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function

$-\int\left(3x^2+4x+2\right)dx$

Expand the integral $\int\left(3x^2+4x+2\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

$-\int3x^2dx-\int4xdx-\int2dx$

The integral of a constant is equal to the constant times the integral's variable

$-\int3x^2dx-\int4xdx-2x$

The integral of a function times a constant ($3$) is equal to the constant times the integral of the function

$-3\int x^2dx-\int4xdx-2x$

The integral of a function times a constant ($4$) is equal to the constant times the integral of the function

$-3\int x^2dx-4\int xdx-2x$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

$-x^{3}-4\int xdx-2x$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$-x^{3}-2x^2-2x$

Since $y$ is treated as a constant, we add a function of $y$ as constant of integration

$-x^{3}-2x^2-2x+g(y)$
4

Integrate $M(x,y)$ with respect to $x$ to get

$-x^{3}-2x^2-2x+g(y)$

The derivative of the constant function ($-x^{3}-2x^2-2x$) is equal to zero

0

The derivative of $g(y)$ is $g'(y)$

$0+g'(y)$
5

Now take the partial derivative of $-x^{3}-2x^2-2x$ with respect to $y$ to get

$0+g'(y)$

Simplify and isolate $g'(y)$

$2\left(y+1\right)=0+g$

$x+0=x$, where $x$ is any expression

$2\left(y+1\right)=g$

Rearrange the equation

$g=2\left(y+1\right)$

Solve the product $2\left(y+1\right)$

$g=2y+2$
6

Set $2\left(y+1\right)$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

$g'(y)=2y+2$

Integrate both sides with respect to $y$

$g=\int\left(2y+2\right)dy$

Expand the integral $\int\left(2y+2\right)dy$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately

$g=\int2ydy+\int2dy$

The integral of a constant is equal to the constant times the integral's variable

$g=\int2ydy+2y$

The integral of a function times a constant ($2$) is equal to the constant times the integral of the function

$g=2\int ydy+2y$

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$g=y^2+2y$
7

Find $g(y)$ integrating both sides

$g(y)=y^2+2y$
8

We have found our $f(x,y)$ and it equals

$f(x,y)=-x^{3}-2x^2-2x+y^2+2y$
9

Then, the solution to the differential equation is

$-x^{3}-2x^2-2x+y^2+2y=C_0$

Group the terms of the equation

$y^2+2y=x^{3}+2x^2+2x+C_0$

Factor the polynomial $y^2+2y$. Add and subtract $\left(\frac{b}{2}\right)^2$, replacing $b$ by it's value $2$

$y^2+2y+1-1=x^{3}+2x^2+2x+C_0$

Now, we can factor $y^2+2x+1$ as a squared binomial of the form $\left(x+\frac{b}{2}\right)^2$

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

We need to isolate the dependent variable $y$, we can do that by subtracting $-1$ from both sides of the equation

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

We can combine and rename $1+C_0$ as other constant of integration

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

Removing the variable's exponent

$y+1=\pm \sqrt{2x^2+x^{3}+2x+C_0}$

We need to isolate the dependent variable $y$, we can do that by subtracting $1$ from both sides of the equation

$y=\pm \sqrt{2x^2+x^{3}+2x+C_0}-1$

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{2x^2+x^{3}+2x+C_0}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

$y=-1+\sqrt{2x^2+x^{3}+2x+C_0},\:y=-1-\sqrt{2x^2+x^{3}+2x+C_0}$
10

Find the explicit solution to the differential equation

$y=-1+\sqrt{2x^2+x^{3}+2x+C_0},\:y=-1-\sqrt{2x^2+x^{3}+2x+C_0}$

$y=-1+\sqrt{2x^2+x^{3}+2x+C_0},\:y=-1-\sqrt{2x^2+x^{3}+2x+C_0}$
SnapXam A2

### beta Got another answer? Verify it!

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1
2
3
4
5
6
7
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

$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$