MHB Equation of tangents and separation of variables question

[ardentmed]

Hey guys,

Can anyone help me out with these questions?

View attachment 2799

The first one has a positive initial value. Separation of variables and integrating gave me: |y+3| = k√[(t^2) + 1)]

Ultimately, I got k= √5 and thus y=√5√[(t^2) + 1)] - 3.

Also, for the second one, I used a similar process, found C=-1/2, and my final answer came out to be:
y=2(x^2) +2. Am I on the right track?

Any help would be much appreciated.

Thanks in advance.

 

[MarkFL]

For the first one, you can drop the absolute value, since your constant will account for all solutions, hence:

$$y(t)=c_1\sqrt{t^2+1}-3$$

Otherwise your work is fine. :D

For the second one, you have made an error. What did you get when you separated the variables?

 

[ardentmed]

I took the integral of 1/(y^2) and 1/(x^3)

-(1/y) = 1(2x^2) + C
C= -1 +1/2
C= -1/2

Thanks.

 

[Prove It]

I took the integral of 1/(y^2) and 1/(x^3)

-(1/y) = 1(2x^2) + C
C= -1 +1/2
C= -1/2

Thanks.

$\displaystyle \begin{align*} \int{ \frac{1}{x^3} \, \mathrm{d}x} &= \int{ x^{-3}\,\mathrm{d}x } \\ &= \frac{x^{-2}}{-2} + C_1 \\ &= - \frac{1}{2x^2} + C_1 \end{align*}$

 

[ardentmed]

$\displaystyle \begin{align*} \int{ \frac{1}{x^3} \, \mathrm{d}x} &= \int{ x^{-3}\,\mathrm{d}x } \\ &= \frac{x^{-2}}{-2} + C_1 \\ &= - \frac{1}{2x^2} + C_1 \end{align*}$

I actually got everything right up to that point. So where did I go wrong? If C= -1/2, then the final answer should be 2x^2 + 2, no?

 

[MarkFL]

I actually got everything right up to that point. So where did I go wrong? If C= -1/2, then the final answer should be 2x^2 + 2, no?

What do you have right after you integrate?

 

[ardentmed]

What do you have right after you integrate?

-1 / y = - 1/(x^2) +C

And then I proceeded to solve for C.

Thanks.

 

[MarkFL]

-1 / y = - 1/(x^2) +C

And then I proceeded to solve for C.

Thanks.

Well, what you should actually have is:

$$-\frac{1}{y}=-\frac{1}{2x^2}+C$$

Now, let's multiply through by -1 and observing that this need not change the sign of the arbitrary constant (a constant times an arbitrary constant is still just an arbitrary constant), we have:

$$\frac{1}{y}=\frac{1}{2x^2}+C$$

Now, solve for $y$, and then use your initial condition to find $C$.

 

[ardentmed]

Well, what you should actually have is:

$$-\frac{1}{y}=-\frac{1}{2x^2}+C$$

Now, let's multiply through by -1 and observing that this need not change the sign of the arbitrary constant (a constant times an arbitrary constant is still just an arbitrary constant), we have:

$$\frac{1}{y}=\frac{1}{2x^2}+C$$

Now, solve for $y$, and then use your initial condition to find $C$.

Oh, the sign doesn't change. Therefore, the final answer should be y= (2x^2) -1, correct? C=-1

 

[MarkFL]

Oh, the sign doesn't change. Therefore, the final answer should be y= (2x^2) -1, correct? C=-1

No, you need to express the right side as a single ratio before inverting both sides:

$$\frac{1}{y}=\frac{1+2Cx^2}{2x^2}$$

Now, $2C$ is just an arbitrary constant, so we may write:

$$\frac{1}{y}=\frac{1+Cx^2}{2x^2}$$

Now, you can flip or invert both sides, and then find $C$. :D

 

[ardentmed]

No, you need to express the right side as a single ratio before inverting both sides:

$$\frac{1}{y}=\frac{1+2Cx^2}{2x^2}$$

Now, $2C$ is just an arbitrary constant, so we may write:

$$\frac{1}{y}=\frac{1+Cx^2}{2x^2}$$

Now, you can flip or invert both sides, and then find $C$. :D

Alright. Therefore, y = (2x^2)/(1/2Cx^2)

1= 2/(1+C), thus making C=1.

So,

y= (2x^2)/(1+x^2)

Correct?

Thanks again.

 

[MarkFL]

Alright. Therefore, y = (2x^2)/(1/2Cx^2)

1= 2/(1+C), thus making C=1.

So,

y= (2x^2)/(1+x^2)

Correct?

Thanks again.

You have a typo in your first line...it should read:

y = (2x^2)/(1+Cx^2)

But other than that, you have the correct solution. Good job! (Yes)