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

3 Answers
Sean
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#

Steve M
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 #

Cesareo R.
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#

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