# How can I solve this differential equation? : #(2x^3-y)dx+xdy=0#

##### 3 Answers

#### Explanation:

.

A first order Ordinary Differential Equation has the form of:

The general solution for this is:

Let's let

Now, we rewrite this in the form of the first order ODE given above. To do so, let's divide

Let's move

Comparing this equation with the general form given above, shows that:

Now, we will find the integrating factor

But we indicated above that

Let's divide both sides by

Since we know that we have the following formula for calculating the derivative of a natural log function:

we can rewrite

We can now take the integral of both sides:

But

Let's substitute this for the left hand side of

We can use the rule of logarithms that says:

to rewrite the right hand side of

Therefore:

This means:

We can test the validity of this answer by plugging it into

Since the whole differential equation will be multiplied by

But we know that if we differentiate

Therefore:

Now, we integrate both sides:

Let's multiply both sides by

# y = Cx - x^3 #

#### Explanation:

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

We have:

# (2x^3-y)dx+xdy = 0 #

Which we can equivalently write in the above standard form as:

# dy/dx - y/x = -2x^2 # ..... [A]

So we compute and integrating factor,

# I = e^(int P(x) dx) #

# \ \ = exp(int \ -1/x \ dx) #

# \ \ = exp( -lnx) #

# \ \ = 1/x #

And if we multiply the DE [A] by this Integrating Factor,

# 1/xdy/dx - y/x^2 = -2x #

# :. d/dx( y/x) = -2x #

This has transformed our initial ODE into a Separable ODE, so we can now *"separate the variables"* to get::

# y/x = int \ -2x \ dx #

This is a standradfunction, so we can integrate to get:

# y/x = -x^2 + C #

Leading to the General Solution of the ODE:

# y = Cx - x^3 #

See below.

#### Explanation:

Now dividing by