# Step-by-step Solution

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

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### Videos

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

## Step-by-step explanation

Problem to solve:

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

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

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

Simplify the expression $2\left(y+1\right)dy$

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

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

$\int2\left(y+1\right)dy=\int\left(3x^2+4x+2\right)dx$
4

The integral of the sum of two or more functions is equal to the sum of their integrals

$\int2\left(y+1\right)dy=\int3x^2dx+\int\left(4x+2\right)dx$
5

The integral of the sum of two or more functions is equal to the sum of their integrals

$\int2\left(y+1\right)dy=\int3x^2dx+\int4xdx+\int2dx$
6

Solve the integral $\int2\left(y+1\right)dy$ and replace the result in the differential equation

$y\left(y+2\right)=\int3x^2dx+\int4xdx+\int2dx$
7

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

$y\left(y+2\right)=\int3x^2dx+\int4xdx+2x$
8

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$y\left(y+2\right)=3\int x^2dx+\int4xdx+2x$
9

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$y\left(y+2\right)=3\int x^2dx+4\int xdx+2x$
10

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 $2$

$y\left(y+2\right)=3\left(\frac{x^{3}}{3}\right)+4\int xdx+2x$
11

Simplify the fraction $3\left(\frac{x^{3}}{3}\right)$

$y\left(y+2\right)=x^{3}+4\int xdx+2x$
12

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 $n=1$

$y\left(y+2\right)=x^{3}+2x^2+2x$
13

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$y\left(y+2\right)=x^{3}+2x^2+2x+C_0$
14

Factor by the greatest common divisor $2$

$y\left(y+2\right)=x^{3}+2\left(x^2+x\right)+C_0$
15

Factor the polynomial $x^2+x$. Add and subtract $\left(\frac{b}{2}\right)^2$, where in this case $b$ equals $1$

$y\left(y+2\right)=x^{3}+2\left(x^2+x+\frac{1}{4}-\frac{1}{4}\right)+C_0$
16

Now we can factor $x^2+x+\frac{1}{4}$ as a squared binomial of the form $\left(x+\frac{b}{2}\right)^2$

$y\left(y+2\right)=x^{3}+2\left(\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right)+C_0$
17

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

$y\cdot y+2y=x^{3}+2\left(\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right)+C_0$
18

Solve the product $2\left(\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right)$

$y\cdot y+2y=x^{3}+2\left(x+\frac{1}{2}\right)^2-\frac{1}{2}+C_0$
19

When multiplying two powers that have the same base ($y$), you can add the exponents

$y^2+2y=x^{3}+2\left(x+\frac{1}{2}\right)^2-\frac{1}{2}+C_0$
20

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

$y^2+2y+1-1=x^{3}+2\left(x+\frac{1}{2}\right)^2-\frac{1}{2}+C_0$
21

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

$\left(y+1\right)^2-1=x^{3}+2\left(x+\frac{1}{2}\right)^2-\frac{1}{2}+C_0$
22

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

$\left(y+1\right)^2=\frac{1}{2}+x^{3}+2\left(x+\frac{1}{2}\right)^2+C_0$
23

Removing the variable's exponent

$y+1=\pm \sqrt{\frac{1}{2}+x^{3}+2\left(x+\frac{1}{2}\right)^2+C_0}$
24

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

$y=\pm \sqrt{\frac{1}{2}+x^{3}+2\left(x+\frac{1}{2}\right)^2+C_0}-1$
25

As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\sqrt{\frac{1}{2}+x^{3}+2\left(x+\frac{1}{2}\right)^2+C_0}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign

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

$y=-1+\sqrt{\frac{1}{2}+x^{3}+2\left(x+\frac{1}{2}\right)^2+C_0},\:y=-1-\sqrt{\frac{1}{2}+x^{3}+2\left(x+\frac{1}{2}\right)^2+C_0}$
$\frac{dy}{dx}=\frac{3x^2+4x+2}{2\left(y+1\right)}$