What is a solution to the differential equation #dy/dx=y^2/x^2+y/x+1#?

Answer 1

# y = xtan (ln x + C)#

This is a first order linear homogeneous equation. NB here homogeneous has its own meaning. it means that the equation can be written in form #y' = f(x,y) # and that #f(kx, ky) = f(x,y)# for constant k.
so standard approach is to let #v = y/x#
so #y = v * x# #y' = v' x + v#

thus, plugging this into the original.....

#v' x + v = v^2 + v + 1#
#v' x = v^2 + 1#
#(v')/(v^2 + 1) = 1/(v^2 + 1) (dv)/dx = 1/x#
#int (dv)/(v^2 + 1) =int 1/x \ dx#
standard integral: #tan^(-1) v = ln x + C#
#tan^(-1) (y/x) = ln x + C#
# (y/x) = tan (ln x + C)#
# y = xtan (ln x + C)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The differential equation ( \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{y}{x} + 1 ) can be solved using an integrating factor.

First, rewrite the equation in a standard form:

[ \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{y}{x} + 1 ]

[ \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2} + 1 ]

The integrating factor ( \mu(x) ) is given by ( \mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln{x}} = \frac{1}{x} ).

Multiply both sides of the equation by ( \frac{1}{x} ):

[ \frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \frac{y^2}{x^3} + \frac{1}{x} ]

Now, rewrite the left-hand side as the derivative of ( \left( \frac{y}{x} \right) ):

[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{y^2}{x^3} + \frac{1}{x} ]

Integrate both sides with respect to ( x ):

[ \int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int \left( \frac{y^2}{x^3} + \frac{1}{x} \right) dx ]

[ \frac{y}{x} = -\frac{y^2}{2x^2} + \ln{|x|} + C ]

[ y = -\frac{xy^2}{2x^2} + x \ln{|x|} + Cx ]

[ y = -\frac{y^2}{2x} + x \ln{|x|} + Cx ]

This is the general solution to the differential equation.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7