What is the general solution of the differential equation # dy/dx + y = xy^3 #?

Hint: try a substitution # z = 1/y^2#

Answer 1

# y^2 = 2/(2x + 1 + Ae^(2x) )#

We have:

# dy/dx + y = xy^3 # ..... [A]

As suggested we perform a substitution:

# z = 1/y^2 iff y^2 = 1/z#
The differentiating wrt #x# we have:
# (dz)/(dy) = -2/y^3 => (dy)/(dz) =-y^3/2#

And from the chain rule, we have:

# dy/dx = dy/dz * dz/dx = -y^3/2 dz/dx#

Substituting into the initial Differential Equation [A], we have:

# -y^3/2 dz/dx + y = xy^3 #
Dividing by #y^3# we have:
# -1/2 dz/dx + 1/y^2 = x #
# :. -1/2 dz/dx + 1/(1/z) = x #
# :. dz/dx -2z = -2x # ..... [B}

This substitution has reduced the equation [A] to a Ordinary Differential Equation of the form, which can be solved using an integrating Factor;

# dz/dx + P(x)z=Q(x) #

Then the integrating factor is given by;

# I = e^(int P(x) dx) # # \ \ = exp(int \ -2 \ dx) # # \ \ = exp( -2x ) # # \ \ = e^(-2x) #
And if we multiply the DE [B] by this Integrating Factor, #I#, we will have a perfect product differential;
# dz/dx e^(-2x) -2e^(-2x) z= -2xe^(-2x) # # :. d/dx( e^(-2x)z) = -2xe^(-2x) #

We can now "separate the variables", to get:

# e^(-2x)z = -2 \ int \ xe^(-2x) \ dx # ..... [C]

To integrate this integral we will require an application of Integration By Parts:

Let # { (u,=x, => (du)/dx,=1), ((dv)/dx,=e^(-2x), => v,=-1/2e^(-2x) ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

gives us

# int \ (x)(e^(-2x)) \ dx = (x)(-1/2e^(-2x)) - int \ (-1/2e^(-2x))(1) \ dx # # :. int \ xe^(-2x) \ dx = -1/2 xe^(-2x) + 1/2 int \ e^(-2x) \ dx # # :. int \ xe^(-2x) \ dx = -1/2 xe^(-2x) -1/4 e^(-2x) #

Using this result, we can now integrate [C] to get:

# e^(-2x)z = -2 {-1/2 xe^(-2x) -1/4 e^(-2x) } +c #
# :. ze^(-2x) = xe^(-2x) +1/2 e^(-2x) +c #
# :. z = x +1/2 +c e^(2x)#

Restoring the substitution we get:

# :. 1/y^2 = 1/2(2x +1 +2ce^(2x) )#
# :. y^2 = 2/(2x +1 +2ce^(2x) )#

Which we can write as:

# y^2 = 2/(2x + 1 + Ae^(2x) )#

Which is the general solution, or:

# y = +-sqrt(2/(2x + 1 + Ae^(2x) ))#
The initial conditions are invalid as #y# is given as a function, rather than a value!
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Answer 2

The general solution of the given differential equation ( \frac{dy}{dx} + y = xy^3 ) is:

[ y(x) = \frac{1}{\sqrt{c - x^2 + c x^4}} ]

where ( c ) is a constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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