Find the general solution of the differential equation

In summary, the conversation discusses solving a differential equation using the method of separation of variables. The resulting equation is solved for x using the initial condition, but the process becomes tricky due to the undefined x'(0). The solution is x = \pm\sqrt{\frac{2}{3}t^3+2t} -2 with a constant of zero.
  • #1
Umayer
13
0

Homework Statement



The equation:

[tex]\frac{dx}{dt}=\frac{t^2+1}{x+2}.[/tex]

Where the initial value is: x(0) = -2.

Homework Equations



I believe you have to use the method of seperations of variables.

The Attempt at a Solution



So I multiplied both sides with x+2. Then I integrated both sides with respect to t.

[tex](x+2)\frac{dx}{dt}=t^2+1[/tex]

[tex]\int(x+2)\frac{dx}{dt}\,dt=\int (t^2+1)\,dt[/tex]

[tex]\int(x+2)\,dx=\int (t^2+1)\,dt[/tex]

[tex]\frac{x^2}{2}+2x=\frac{t^3}{3}+t+C[/tex]

(Note: I've added the two constants of both sides into one constant.)

Then, I multiplied everything with 2.

[tex]x^2+4x=\frac{2}{3}t^3+2t+C'[/tex]

Where C'=2C

Now I'm not so sure how I should go further then this. Any help would be nice.
 
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  • #2
It's correct so far -- if you have to solve for x, I think completing the square would work nicely.
 
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  • #3
Use the initial condition to find the constant C'.

ehild
 
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  • #4
Umayer said:

Homework Statement



The equation:

[tex]\frac{dx}{dt}=\frac{t^2+1}{x+2}.[/tex]

Where the initial value is: x(0) = -2.

Homework Equations



I believe you have to use the method of seperations of variables.

The Attempt at a Solution



So I multiplied both sides with x+2. Then I integrated both sides with respect to t.

[tex](x+2)\frac{dx}{dt}=t^2+1[/tex]

[tex]\int(x+2)\frac{dx}{dt}\,dt=\int (t^2+1)\,dt[/tex]

[tex]\int(x+2)\,dx=\int (t^2+1)\,dt[/tex]

[tex]\frac{x^2}{2}+2x=\frac{t^3}{3}+t+C[/tex]

(Note: I've added the two constants of both sides into one constant.)

Then, I multiplied everything with 2.

[tex]x^2+4x=\frac{2}{3}t^3+2t+C'[/tex]

Where C'=2C

Now I'm not so sure how I should go further then this. Any help would be nice.

[itex]x^2 + 4x = (x + 2)^2 - 4[/itex].

You might ask "what sign goes before the square root?", and normally the answer would be "use the ODE to work out whether [itex]x'(0)[/itex] is positive or negative", but in this case the ODE tells you that [itex]x'(0)[/itex] is undefined ...
 
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  • #5
Thanks allot guys I've found the answer.
 
  • #6
Umayer said:
Thanks allot guys I've found the answer.

Just as a matter of interest: what is your solution? (This IVP is a bit tricky!)
 
  • #7
Ray Vickson said:
Just as a matter of interest: what is your solution? (This IVP is a bit tricky!)

[tex]x = \pm\sqrt{\frac{2}{3}t^3+2t} -2[/tex]

The constant is zero. It is tricky indeed.
 

FAQ: Find the general solution of the differential equation

1. What is a differential equation?

A differential equation is an equation that involves one or more derivatives of a function. It describes how a particular physical quantity changes over time or in relation to other variables.

2. What is the general solution of a differential equation?

The general solution of a differential equation is a family of functions that satisfies the equation. It contains an arbitrary constant that can take on any value, resulting in a set of solutions.

3. How do you find the general solution of a differential equation?

To find the general solution of a differential equation, you need to integrate the equation using appropriate techniques. This involves finding an antiderivative of the function and applying any initial conditions to determine the value of the arbitrary constant.

4. What is the order of a differential equation?

The order of a differential equation is the highest derivative present in the equation. For example, a first-order differential equation contains only first derivatives, while a second-order differential equation contains second derivatives.

5. What are initial conditions in a differential equation?

Initial conditions are the values of the dependent variable and its derivatives at a specific point in the domain. They are used to determine the value of the arbitrary constant in the general solution of a differential equation.

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