Find the general solution of the differential equation

[Umayer]

Homework Statement

The equation:

\frac{dx}{dt}=\frac{t^2+1}{x+2}.

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.

(x+2)\frac{dx}{dt}=t^2+1

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

\int(x+2)\,dx=\int (t^2+1)\,dt

\frac{x^2}{2}+2x=\frac{t^3}{3}+t+C

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

Then, I multiplied everything with 2.

x^2+4x=\frac{2}{3}t^3+2t+C'

Where C'=2C

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

 

[jackarms]

It's correct so far -- if you have to solve for x, I think completing the square would work nicely.

 

[ehild]

Use the initial condition to find the constant C'.

ehild

 

[pasmith]

Homework Statement

The equation:

\frac{dx}{dt}=\frac{t^2+1}{x+2}.

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.

(x+2)\frac{dx}{dt}=t^2+1

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

\int(x+2)\,dx=\int (t^2+1)\,dt

\frac{x^2}{2}+2x=\frac{t^3}{3}+t+C

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

Then, I multiplied everything with 2.

x^2+4x=\frac{2}{3}t^3+2t+C'

Where C'=2C

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

x^2 + 4x = (x + 2)^2 - 4.

You might ask "what sign goes before the square root?", and normally the answer would be "use the ODE to work out whether x'(0) is positive or negative", but in this case the ODE tells you that x'(0) is undefined ...

 

[Umayer]

Thanks allot guys I've found the answer.

 

[Ray Vickson]

Thanks allot guys I've found the answer.

Just as a matter of interest: what is your solution? (This IVP is a bit tricky!)

 

[Umayer]

Just as a matter of interest: what is your solution? (This IVP is a bit tricky!)

x = \pm\sqrt{\frac{2}{3}t^3+2t} -2

The constant is zero. It is tricky indeed.