Find the general solution of the differential equation

[Umayer]

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.

 

[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:

[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 ...

 

[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!)

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

The constant is zero. It is tricky indeed.