Unseparable Differential Equations

In summary, the conversation is about solving a first order linear differential equation. The initial equation is given as xy' + y = y^2, which is separable and non-linear. The attempt at a solution involves finding the integrating factor and using the general formula for linear first order differential equations. However, the issue arises in isolating for y in the final solution. The conversation then moves on to discussing how to isolate for y and finding the final solution with the given initial condition.
  • #1
tascja
87
0
I have a question about differential equations... The equation for a general differential equation that is not separable is:
dy/dx + P(x)y = Q(x)

So my question is can you have a Q(x) that has both x and y variables?

Example:

Homework Statement


xy' +y = y^2

The Attempt at a Solution


dy/dx + (1/x)y = (1/x)y^2

The integrating factor V(x) = e ^[int. P(x)]
= e ^[int. 1/x dx]
= x

Therefore y = [1/V(x)][int. Q(x)V(x)dx + C]
= [1/x] [int. ? + C]
 
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  • #2
tascja said:
I have a question about differential equations... The equation for a general LINEAR first order differential equation (separable or not) is:
dy/dx + P(x)y = Q(x)

So my question is can you have a Q(x) that has both x and y variables?

Example:

Homework Statement


xy' +y = y^2
This is separable! (and non-linear, by the way) Proceed as you normally would in that case.
 
Last edited:
  • #3


Yes, sorry as i was looking over i finally remembered its called a first order linear differential equation.. sorry :$... but i tried to solve by separating but don't seem to get the right answer.. here's what i got for a solution (with the initial condition being y(1)=-1)

-1/y - ln|y| = ln|x| + 1

but how would i isolate y??
 
  • #4


tascja said:
Yes, sorry as i was looking over i finally remembered its called a first order linear differential equation.. sorry :$... but i tried to solve by separating but don't seem to get the right answer.. here's what i got for a solution (with the initial condition being y(1)=-1)

-1/y - ln|y| = ln|x| + 1

but how would i isolate y??
How did you get that?

We have [tex]\frac{1}{y(y-1)} y' = \frac{1}{x}[/tex]

and the antiderivative of

[tex]\frac{1}{y(y-1)} = \frac{1}{y-1} - \frac{1}{y}[/tex]

is ?
 
  • #5
even still.. i don't see how you can isolate for y without y in the answer??
 
  • #6
tascja said:
even still.. i don't see how you can isolate for y without y in the answer??

Why would you need to isolate y? If you were given y(1)=-1, that means that when x=1,y=-1. No need to isolate y.
 
  • #7
tascja said:
even still.. i don't see how you can isolate for y without y in the answer??
You didn't answer my question. Without seeing your response I can't know why you can't isolate y.
 
  • #8


oh sorry...
so since you said
Unco said:
How did you get that?

We have [tex]\frac{1}{y(y-1)} y' = \frac{1}{x}[/tex]

and the antiderivative of

[tex]\frac{1}{y(y-1)} = \frac{1}{y-1} - \frac{1}{y}[/tex]

is ?
therefore having taking the antiderivative of both sides,
the next line would be:
1/y-1 - 1/y = ln|x| + C
then subbing in y=-1 and x=1 to solve for C:
1/1 +1/1 = ln|1| + C
C = 2
so
1/y-1 - 1/y = ln|x| + 2 (sorry i think i might have put 1 before as C)
so now i need to enter an answer that is y=
and i don't know how to isolate for y??
 
  • #9


tascja said:
oh sorry...
so since you said

therefore having taking the antiderivative of both sides,
the next line would be:
1/y-1 - 1/y = ln|x| + C
then subbing in y=-1 and x=1 to solve for C:
1/1 +1/1 = ln|1| + C
C = 2
so
1/y-1 - 1/y = ln|x| + 2 (sorry i think i might have put 1 before as C)
so now i need to enter an answer that is y=
and i don't know how to isolate for y??
You still didn't answer my question, tascja!

The antiderivative of 1/(y-1) - 1/y is ln(y-1) - ln(y) = ln((y-1)/y) ...
 
  • #10
omg i thought you were implying that 1/(y-1) - 1/y was the antiderivative of 1/y(y-1) that's why i was confused, sorry.. again... okay so
ln((y-1)/y)) = ln|x| +c
(y-1)/y = e^[ln|x| +c]
1- 1/y = e^[ln|x|]e^C *e^C is just arbitrary number
y = 1/(1-Cx) * where with the initial conditions C=2
y = 1/(1-2x)

THANK YOU FOR ALL YOUR HELP! :D
 
  • #11
tascja said:
omg i thought you were implying that 1/(y-1) - 1/y was the antiderivative of 1/y(y-1) that's why i was confused, sorry.. again... okay so
ln((y-1)/y)) = ln|x| +c
(y-1)/y = e^[ln|x| +c]
1- 1/y = e^[ln|x|]e^C *e^C is just arbitrary number
y = 1/(1-Cx) * where with the initial conditions C=2
y = 1/(1-2x)
If what I wrote was unclear it is I who should be apologising! Well done!
 

FAQ: Unseparable Differential Equations

1. What are unseparable differential equations?

Unseparable differential equations are a type of differential equation where the dependent variable and the independent variable cannot be separated on each side of the equation. This means that the equation cannot be solved using traditional algebraic methods.

2. How do you solve unseparable differential equations?

Unseparable differential equations can be solved using various techniques such as finding an integrating factor, using substitution, or solving as a power series. The method used will depend on the specific equation and its complexity.

3. Why are unseparable differential equations important?

Unseparable differential equations are important in many fields of science and engineering, as they allow us to model and understand various physical phenomena. They are also essential in solving real-world problems and making predictions about the behavior of systems.

4. Can unseparable differential equations have multiple solutions?

Yes, unseparable differential equations can have multiple solutions. This is because the equations may have different initial conditions or constants that result in different solutions. It is important to carefully consider the initial conditions and any given constraints when solving these equations.

5. How can I check if my solution to an unseparable differential equation is correct?

The best way to check if your solution to an unseparable differential equation is correct is by substituting the solution back into the original equation and seeing if it satisfies the equation. You can also use numerical methods to approximate the solution and compare it to your analytical solution.

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