Solving ODEs Passing Through Points: x'=x^{\frac{1}{2}}

[fluidistic]

Homework Statement

1)Find the solution of $$x'=x^{\frac{1}{2}}$$ that passes through the point $$(t_0, x_0)$$ where $$x_0>0$$.
2)Find all the solutions of this equation that pass through the point $$(t_0,0)$$.

Homework Equations

Direct integration.

The Attempt at a Solution

1)$$\frac{dx}{dt}=x^{\frac{1}{2}} \Rightarrow \frac{dx}{x^{\frac{1}{2}}}=dt \Rightarrow \int \frac{dx}{x^{\frac{1}{2}}}=t+C\Rightarrow 2 x^{\frac{1}{2}}=t+C \Rightarrow x=\frac{t^2}{4}+tC+C^2$$.

I determined C thanks to the initial condition and the equation became $$x=\frac {t^2}{4}+t \left ( \frac{2x_0 ^{\frac{1}{2}}-t_0}{2}} \right ) + \frac{(2x_0 ^{\frac{1}{2}}-t_0)^2}{4}}$$.
2) Replacing $$x_0$$ by $$0$$ in the above equation yields $$x= \left ( \frac{t}{2}-\frac{t_0}{2} \right ) ^2$$.
Unfortunately I replaced this solution into the original equation and the equality isn't true. So I made an error. I also replaced the first solution I got (the one with C's) into the equation and it didn't work. Hence I made an error quite early. I don't know where though.

 

[tiny-tim]

Hi fluidistic!

1)$$\cdots 2 x^{\frac{1}{2}}=t+C \Rightarrow x=\frac{t^2}{4}+tC+C^2$$.

Nooo

 

[fluidistic]

Hi fluidistic!


Nooo

I got it thanks. Yeah a shame!

 

[cronxeh]

Why do people have such an aversion to google or wolfram alpha?

http://www.wolframalpha.com/input/?i=dx/dt+=+sqrt(x)"

http://www.wolframalpha.com/input/?i=dx/dt+=+sqrt(x),+x(0)=0"

 

[fluidistic]

Why do people have such an aversion to google or wolfram alpha?

http://www.wolframalpha.com/input/?i=dx/dt+=+sqrt(x)"

http://www.wolframalpha.com/input/?i=dx/dt+=+sqrt(x),+x(0)=0"

I don't hate those. I knew I was wrong as I stated in my first post. I didn't know where though. Wolfram alpha would have showed that I was wrong early, but this, I already knew. Thanks anyway.

 

[cronxeh]

I don't hate those. I knew I was wrong as I stated in my first post. I didn't know where though. Wolfram alpha would have showed that I was wrong early, but this, I already knew. Thanks anyway.

Please post your final solution. There are some mistakes that make me go

 

[fluidistic]

Please post your final solution. There are some mistakes that make me go

Ok. I have at least one doubt.
Here goes my attempt: $$x=\frac{t^2+2tC+C^2}{4}$$. It satisfies the DE, so I'm sure I'm right until here.
Now to answer 1): $$x_0=x(t_0)=\frac{t_0^2+2t_0C+C^2}{4}$$.
Thus $$C^2+2t_0C+(t_0^2-4x_0)=0$$. Solving for C, I got $$C=-t_0 \pm 2 x_0^{\frac{1}{2}}$$.
Hence $$x= \frac{t^2+2t (2x_0^ {\frac{1}{2}} -t_0)+(2x_0^{\frac{1}{2}}-t_0 )^2}{4}$$.
Notice that I took $$C=-t_0+2 x_0^{\frac{1}{2}}$$ but I'm not sure why I didn't take $$C=-t_0-2 x_0^{\frac{1}{2}}$$. This is my doubt.
To answer 2), I just plugged $$x_0=0$$ in the above equation to reach oh... yeah the same wrong result I had.
I'm really clueless where I made an error. I revised the quadratic in C and I don't see any error. The first equation is right... so, maybe my method isn't. I'd love to know where are my mistakes.

 

[cronxeh]

Ok. I have at least one doubt.
Here goes my attempt: $$x=\frac{t^2+2tC+C^2}{4}$$. It satisfies the DE, so I'm sure I'm right until here.
Now to answer 1): $$x_0=x(t_0)=\frac{t_0^2+2t_0C+C^2}{4}$$.
Thus $$C^2+2t_0C+(t_0^2-4x_0)=0$$. Solving for C, I got $$C=-t_0 \pm 2 x_0^{\frac{1}{2}}$$.
Hence $$x= \frac{t^2+2t (2x_0^ {\frac{1}{2}} -t_0)+(2x_0^{\frac{1}{2}}-t_0 )^2}{4}$$.
Notice that I took $$C=-t_0+2 x_0^{\frac{1}{2}}$$ but I'm not sure why I didn't take $$C=-t_0-2 x_0^{\frac{1}{2}}$$. This is my doubt.
To answer 2), I just plugged $$x_0=0$$ in the above equation to reach oh... yeah the same wrong result I had.
I'm really clueless where I made an error. I revised the quadratic in C and I don't see any error. The first equation is right... so, maybe my method isn't. I'd love to know where are my mistakes.

Ok let's rewrite it as:

x(t) = 1/4 (2*C*t + C^2 + t^2)

x_0 means x(0) means t=0

x(0) = 1/4*C^2 C = sqrt(4*x0) = 2*sqrt(x0)

So for problem #1 you are asked to find x(t) = 1/4* (2*2*sqrt(x0)*t + 4*x0 + t^2)

and x(t) = 1/4*(4*sqrt(x0)*t + 4*x0 + t^2)

for problem #2 you are asked to find solutions that pass through (t0,x0=0)

x(t) = 1/4*(4*sqrt(0)+4*0+t^2) = 1/4 * t^2 = (t^2)/4

 

[fluidistic]

Ok let's rewrite it as:

x(t) = 1/4 (2*C*t + C^2 + t^2)

x_0 means x(0) means t=0

x(0) = 1/4*C^2 C = sqrt(4*x0) = 2*sqrt(x0)

So for problem #1 you are asked to find x(t) = 1/4* (2*2*sqrt(x0)*t + 4*x0 + t^2)

and x(t) = 1/4*(4*sqrt(x0)*t + 4*x0 + t^2)

for problem #2 you are asked to find solutions that pass through (t0,x0=0)

x(t) = 1/4*(4*sqrt(0)+4*0+t^2) = 1/4 * t^2 = (t^2)/4

Thanks a lot for the answer, I'm going back to it as soon as I can, i.e. in about 2 days... I have to do a trip of more than 800 km (11 hours) due to a passport thing.

 

[fluidistic]

I'm back!
Ok, it seems much easier than I thought. But I have some questions, why did you chose $$C=2x_0 ^{\frac{1}{2}}$$ and not $$C=-2x_0 ^{\frac{1}{2}}$$?
My other questions are that although I misunderstood what was meant by $$x_0$$ (I thought it was x(t_0) instead of x(0) ), why won't my final expression of x(t) work? I should have found a particular solution to the equation... or not?
Also, what's the deal with this $$t_0$$? It doesn't even appear a single time anywhere but in the question.

 

[fluidistic]

Small bump... can someone answer my last 3 questions? I'm self studying DE's, I really need to understand the topic and my 3 last questions are puzzling me.
Thanks in advance for any help!