# First order DE

Gold Member

## 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
Homework Helper
Hi fluidistic!
1)$$\cdots 2 x^{\frac{1}{2}}=t+C \Rightarrow x=\frac{t^2}{4}+tC+C^2$$.

Nooo

Gold Member
Hi fluidistic!

Nooo

I got it thanks. Yeah a shame!

cronxeh
Gold Member
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

Gold Member
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
Gold Member
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 lets 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

Gold Member
Ok lets 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.

Gold Member
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.

Gold Member
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!