Manipulation of Arbitrary Constants in Differential Equations

[Bill Nye Tho]

Homework Statement

yy''+(y')² = 0

Homework Equations

yv(dv/dy)+v²=0

The Attempt at a Solution

Variable separable when solving for the first step the result is:

- ln |v| = ln |y| + C₁

Now, I've looked at the remainder of the solution with a few other sources and the cause of my mistake results in the constant.

After turning the equation into: ln |vy| + C₁ = 0;

I raise everything to the e so that I can solve for v.

It seems all the solutions do that as well but yet they don't raise the constant to the e.

Is it because e^C₁ will still be a constant and therefore completely arbitrary?

 

[SammyS]

Homework Statement

yy''+(y')² = 0

Homework Equations

yv(dv/dy)+v²=0

The Attempt at a Solution

Variable separable when solving for the first step the result is:

- ln |v| = ln |y| + C₁

Now, I've looked at the remainder of the solution with a few other sources and the cause of my mistake results in the constant.

After turning the equation into: ln |vy| + C₁ = 0;

I raise everything to the e so that I can solve for v.

It seems all the solutions do that as well but yet they don't raise the constant to the e.

Is it because e^C₁ will still be a constant and therefore completely arbitrary?

[itex]\displaystyle e^{C_{\,1}}\ [/itex] is positive, otherwise it's as arbitrary as using a redefined C₁ .

 

[HallsofIvy]

Homework Statement

yy''+(y')² = 0

Homework Equations

yv(dv/dy)+v²=0

The Attempt at a Solution

Variable separable when solving for the first step the result is:
-ln|v|= ln|y|+ C₁

You could, at this point, take the exponential of both sides:
[itex]e^{-ln|v|}= e^{ln|y|+ C_1}[/itex]
[itex]\frac{1}{|v|}= e^{C_1}|y|[/itex]
Now, as you say, since C₁ is an arbitrary constant, so is [itex]e^{C_1}[/itex] so let's just call it "C₂". And if allow C₂ to be either positive or negative, it we can drop the absolute values: [itex]\frac{1}{v}= C_2y[/itex] which is, of course, the same as [itex]yv= \frac{1}{C_2}[/itex] or [itex]yv= C_3[/itex] where [itex]C_3= 1/C_2[/itex].
Some texts just don't bother to label the different constants differently.

Now, I've looked at the remainder of the solution with a few other sources and the cause of my mistake results in the constant.

After turning the equation into: ln |vy| + C₁ = 0;

I raise everything to the e so that I can solve for v.

It seems all the solutions do that as well but yet they don't raise the constant to the e.

Is it because e^C₁ will still be a constant and therefore completely arbitrary?

 

[Bill Nye Tho]

You could, at this point, take the exponential of both sides:
[itex]e^{-ln|v|}= e^{ln|y|+ C_1}[/itex]
[itex]\frac{1}{|v|}= e^{C_1}|y|[/itex]
Now, as you say, since C₁ is an arbitrary constant, so is [itex]e^{C_1}[/itex] so let's just call it "C₂". And if allow C₂ to be either positive or negative, it we can drop the absolute values: [itex]\frac{1}{v}= C_2y[/itex] which is, of course, the same as [itex]yv= \frac{1}{C_2}[/itex] or [itex]yv= C_3[/itex] where [itex]C_3= 1/C_2[/itex].
Some texts just don't bother to label the different constants differently.

That makes sense, thank you.

Seems like they just don't want things to get messy so they reuse the same constants.

 

[HallsofIvy]

A "conservation of constants" law!