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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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Final Answer

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

Problem to solve:

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

Specify 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 $f(x,y)=C$

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

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

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

Final Answer

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

Tips on how to improve your answer:

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

Time to solve it:

~ 0.15 s