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

3 Answers
Feb 17, 2018

y=-x^3+xC_1

Explanation:

.

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

A first order Ordinary Differential Equation has the form of:

y'(x)+p(x)y=q(x)

The general solution for this is:

y(x)=(inte^(intp(x)dx)q(x)dx+C)/e^(int(p(x)dx

Let's let y be the dependent variable and divide the equation by dx:

2x^3-y+xd/dx(y)=0 color(red)(Equation 1)

Now, we rewrite this in the form of the first order ODE given above. To do so, let's divide color(red)(Equation 1) by x:

2x^2-1/xy+d/dxy=0

Let's move 2x^2 to the other side and switch the locations of the two remaining terms on the left hand side:

d/dxy-1/xy=-2x^2 color(blue)(ODE)

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

p(x)=-1/x and q(x)=-2x^2

Now, we will find the integrating factor mu(x) such that:

mu(x)*p(x)=mu'(x)

But we indicated above that p(x)=-1/x and mu'(x) is the same as d/dx(mu(x)). Let's plug these in:

mu(x)*(-1/x)=d/dx(mu(x))

Let's divide both sides by mu(x):

(mu(x)*(-1/x))/(mu(x))=(d/dx(mu(x)))/(mu(x))

(cancelcolor(red)(mu(x))*(-1/x))/(cancelcolor(red)(mu(x)))=(d/dx(mu(x)))/(mu(x))

-1/x=(d/dx(mu(x)))/(mu(x)) color(red)(Equation 2)

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

d/dx(ln(f(x)))=(d/dx(f(x)))/(f(x))

we can rewrite color(red)(Equation 2) as:

-1/x=d/dx(ln(mu(x)))

We can now take the integral of both sides:

int-1/xdx=ln(mu(x)) color(red)(Equation 3)

But int-1/xdx=-ln(x)+C_1

Let's substitute this for the left hand side of color(red)(Equation 3):

ln(mu(x))=-ln(x)+C_1 color(red)(Equation 4)

We can use the rule of logarithms that says:

a=log_b(b^a)

to rewrite the right hand side of color(red)(Equation 4):

-ln(x)+C_1=ln(e^(-ln(x)+C_1))=ln((e^(C_1))/e^(ln(x)))=ln(e^(C_1)/x)

Therefore:

ln(mu(x))=ln(e^(C_1)/x)

This means:

mu(x)=e^(C_1)/x

We can test the validity of this answer by plugging it into color(red)(Equation 4) and see that it works.

Since the whole differential equation will be multiplied by e^(C_1)/x, the constant part e^(C_1) can be ignored and:

mu(x)=1/x is our integration factor. Let's multiply the above color(blue)(ODE) by 1/x:

1/x*d/dxy-1/x*1/xy=-1/x*2x^2

(d/dx(y))/x-y/x^2=-2x

But we know that if we differentiate 1/xy using the product rule we get:

d/dx(1/xy)=1/xy'-1/x^2y=(d/dx(y))/x-y/x^2

Therefore:

d/dx(1/xy)=-2x

Now, we integrate both sides:

1/xy=int-2xdx

1/xy=-x^2+C_1

Let's multiply both sides by x:

x*1/xy=x(-x^2)+xC_1

cancelcolor(red)x*1/cancelcolor(red)xy=x(-x^2)+xC_1

y=-x^3+xC_1

Feb 17, 2018

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, using;

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, I, we will have a perfect product differential;

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

Feb 17, 2018

See below.

Explanation:

(2x^3-y)dx+x dy =0 or

2x^3 dx -y dx + x dy = 0

Now dividing by x^2

2x-y/x^2 dx+dy/x = 0 but this is the differential

d(x^2 + y/x) = 0 hence

x^2+y/x=C_0 or

y = C_0 x - x^3