Constant in differential equations

[dunkelheit]

I have a doubt with the use of the additive constants in differential equations, more in particular about the algebra with them.
If we consider the Cauchy's problem

[MATH]\begin{cases} y'(x)=\frac{2}{(y(x))^2} \\ y(0)=1 \end{cases}[/MATH]​

We have that

[MATH]y'(x)(y(x))^2=2\Rightarrow \frac{1}{3}\frac{\text{d}}{\text{d}x} (y(x))^3=2[/MATH]​

Now I have I doubt: if I proceed directly with integration I get that

[MATH]\frac{1}{3} (y(x))^3=2x+c \Rightarrow(y(x))^3=6x+c\Rightarrow y(x)=(6x+c)^{\frac{1}{3}}[/MATH]​

So applying the initial condition I get that [MATH]c=1[/MATH], so the solution of the Cauchy's problem is [MATH]y(x)=(6x+1)^{\frac{1}{3}}[/MATH]; but if when multiplying both sides for [MATH]\frac{1}{3}[/MATH] I multiply the constant [MATH]c[/MATH] too for [MATH]\frac{1}{3}[/MATH] I would get

[MATH]\frac{1}{3} (y(x))^3=2x+c \Rightarrow(y(x))^3=6x+3c\Rightarrow y(x)=(6x+3c)^{\frac{1}{3}}[/MATH]​

So applying the initial condition I would get that [MATH]c=\frac{1}{3}[/MATH], so the solution of the Cauchy's problem is [MATH]y(x)=\left(6x+\frac{1}{3}\right)^{\frac{1}{3}}[/MATH] which is different from the other one. Where is the mistake? Is this because [MATH]2c[/MATH] is another constand and even if [MATH]c[/MATH] is arbitrary I should rename it something like [MATH]\tilde{c}=2c[/MATH]? But then I don't get why it is arbitrary if it changes basing on the algebra we do with it.
I think they are both correct but I don't understand why this works, since [MATH]3c[/MATH] is different from [MATH]c[/MATH].
Thanks.

 

[Steven G]

In 6x+3c you discovered that 3c = 1 so 6x+3c becomes 6x+1

In 6x+c you discovered that c=1 so 6x+c becomes 6x+1.

Where is the trouble?? The two c's are different!

 

[HallsofIvy]

I have a doubt with the use of the additive constants in differential equations, more in particular about the algebra with them.
If we consider the Cauchy's problem

[MATH]\begin{cases} y'(x)=\frac{2}{(y(x))^2} \\ y(0)=1 \end{cases}[/MATH]​

We have that

[MATH]y'(x)(y(x))^2=2\Rightarrow \frac{1}{3}\frac{\text{d}}{\text{d}x} (y(x))^3=2[/MATH]​

Now I have I doubt: if I proceed directly with integration I get that

[MATH]\frac{1}{3} (y(x))^3=2x+c \Rightarrow(y(x))^3=6x+c\Rightarrow y(x)=(6x+c)^{\frac{1}{3}}[/MATH]​

So applying the initial condition I get that [MATH]c=1[/MATH], so the solution of the Cauchy's problem is [MATH]y(x)=(6x+1)^{\frac{1}{3}}[/MATH]; but if when multiplying both sides for [MATH]\frac{1}{3}[/MATH] I multiply the constant [MATH]c[/MATH] too for [MATH]\frac{1}{3}[/MATH] I would get

[MATH]\frac{1}{3} (y(x))^3=2x+c \Rightarrow(y(x))^3=6x+3c\Rightarrow y(x)=(6x+3c)^{\frac{1}{3}}[/MATH]​

So applying the initial condition I would get that [MATH]c=\frac{1}{3}[/MATH], so the solution of the Cauchy's problem is [MATH]y(x)=\left(6x+\frac{1}{3}\right)^{\frac{1}{3}}[/MATH] which is different from the other one. Where is the mistake?

Your mistake is that your solution, before you found a value for c, was
[MATH]\frac{1}{3} (y(x))^3=2x+c \Rightarrow(y(x))^3=6x+3c\Rightarrow y(x)=(6x+3c)^{\frac{1}{3}}[/MATH]
So now that you have determined that c= 1/3, you should have
[math]y(x)= (6x+ 3(1/3))^{\frac{1}{3}}= (6x+ 1)^{\frac{1}{3}}[/math]just as before.

Is this because [MATH]2c[/MATH] is another constand and even if [MATH]c[/MATH] is arbitrary I should rename it something like [MATH]\tilde{c}=2c[/MATH]? But then I don't get why it is arbitrary if it changes basing on the algebra we do with it.
I think they are both correct but I don't understand why this works, since [MATH]3c[/MATH] is different from [MATH]c[/MATH].
Thanks.

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