First order differential equations: (1/x)(dy/dx) = 1/(1 - x^2), (x - 3)(dy/dx) = y, tan(y)(dy/dx) - 4

[Aminta_1900]

I'm approaching this subject for the first time and just need a few hints. I've been getting many questions wrong from what looks like using naive solutions, but the given answers seem to be using solutions I am not familiar with. I'm not sure why my textbook has all of a sudden assumed I should known them!

Here is an example of three questions, with my naive answer and, where they differ from mine, the answers given from symbolab (in blue) and the answers given at the back of my textbook (in red).


Can anyone see where i am generally going wrong, or perhaps show me how natural log and e have been produced in these answers? Many thanks.

 

[Dr.Peterson]

I'm approaching this subject for the first time and just need a few hints. I've been getting many questions wrong from what looks like using naive solutions, but the given answers seem to be using solutions I am not familiar with. I'm not sure why my textbook has all of a sudden assumed I should known them!

Here is an example of three questions, with my naive answer and, where they differ from mine, the answers given from symbolab (in blue) and the answers given at the back of my textbook (in red).

View attachment 36646
Can anyone see where i am generally going wrong, or perhaps show me how natural log and e have been produced in these answers? Many thanks.

It isn't that your work is naive; you've just made little mistakes, and they've done extra simplification (which isn't necessary - and I wish textbooks didn't do that).

1) Your answer is identical to theirs. Try condensing your log (or expanding theirs), and rename the constant. But your answer would be accepted.

2) How did you get y inside your expression for y? Please explain the last step; what you should have done there is to raise e to the power on each side. (The two correct answers are equivalent, just using different names for the constant; and they are equivalent to your next-to-last step.)

3) Where you meant to integrate the cotangent, you differentiated it. Try the integral again!

 

[Aminta_1900]

It isn't that your work is naive; you've just made little mistakes, and they've done extra simplification (which isn't necessary - and I wish textbooks didn't do that).

1) Your answer is identical to theirs. Try condensing your log (or expanding theirs), and rename the constant. But your answer would be accepted.

2) How did you get y inside your expression for y? Please explain the last step; what you should have done there is to raise e to the power on each side. (The two correct answers are equivalent, just using different names for the constant; and they are equivalent to your next-to-last step.)

3) Where you meant to integrate the cotangent, you differentiated it. Try the integral again!

Thanks a lot for the help.

1) I think I have this one now.


2) Here with some expanded workings.....though I'm not that confident with rearranging log and e in formulas, and haven't had much practice, so I'm still a bit stuck here....


3) I think I have this one one as well.....but I'm taking it that the +C can be added to either side as the last step(?) Otherwise I wouldn't know how -C changed to +C.



Thanks again.

 

[khansaheb]

For #2 , you got, ..................$\displaystyle \ln |\frac{y}{x-3}| \ = \ C_1$.............use definition of 'log' .............. log(a) = b .......... implies ......e^b = a

Then you get $\displaystyle |\frac{y}{x-3}| \ = \ e^{C_1}$.....or.... $\displaystyle y = e^{C_1} * (x-3)$........or....... y = C * (x-3) ....where $\displaystyle e^{C_1}$= C

for #3, C is a constant of integration. Thus you could express it (the constant of integration) as "-C" in the last step of integration.

 

[Dr.Peterson]

1) I think I have this one now.

As I said before, what you wrote last time would be an acceptable answer; here you've just gone part of the way to the "simplified" form, by condensing the log. Renaming C as $\ln|C|$ is useful only if you make the next step, combining the two logs using the fact that $ln(A)+ln(B)=ln(AB)$.

2) Here with some expanded workings.....though I'm not that confident with rearranging log and e in formulas, and haven't had much practice, so I'm still a bit stuck here....

From $\ln|y|-\ln|x-3|=C$, you first rearranged as $\ln|y|=\ln|x-3|+C$, which is good. But to solve for y, you need to exponentiate each side like this: $e^{\ln|y|}=e^{\ln|x-3|+C}$. Then you can use the fact that $e^{\ln(A)}=A$ to get y by itself.

3) I think I have this one one as well.....but I'm taking it that the +C can be added to either side as the last step(?) Otherwise I wouldn't know how -C changed to +C.

If you are not required to solve for C, then this is valid. As for C changing sign, that's a matter of what I called "renaming C". Since C is just an arbitrary constant, you can rename $-C$ as $C_1$, and then, at the end, just call that $C$.

You need to go back to your algebra and master logarithms!

 

[Dr.Peterson]

If you are not required to solve for C, then this is valid.

I meant, "If you are not required to solve for y, then this is valid."

3) I think I have this one one as well.....but I'm taking it that the +C can be added to either side as the last step(?) Otherwise I wouldn't know how -C changed to +C.

Let's fill in a gap or two, and then solve for y, as you showed Symbolab did.

First, what you did (aiming at your book's answer, apparently) was to multiply by 4 and move C:
$$\frac{1}{4}x=\ln|\sin y|+C\\x=4\ln|\sin y|+4C\\x-4C=4\ln|\sin y|\\x+C_1=4\ln|\sin y|$$
At the last step, I set $C_1 = -4C$, which is just a different arbitrary constant.

To solve for y, we can now divide by 4 again, and exponentiate:
$$\frac{x+C_1}{4}=\ln|\sin y|\\e^{(x+C_1)/4}=e^{\ln|\sin y|}\\e^{(x+C_1)/4}=|\sin y|\\\sin y=\pm e^{(x+C_1)/4}$$
We could then use the inverse sine to solve. But Symbolab apparently prefers to use the inverse cosine, probably because it makes it a little easier to show all solutions. That's not something I would have thought of, and I don't at a glance see how they got $4x+1$; but maybe we can do that using the fact that $\cos^2(x)=1-\sin^2(x)$. Let's see:
$$\sin^2y=e^{2(x+C_1)/4}\\1-\sin^2y=1-e^{(x+C_1)/2}\\\cos^2y=1-e^{(x+C_1)/2}\\\cos y=\pm\left(1-e^{(x+C_1)/2}\right)^{1/2}$$That doesn't work.

Actually, when I graph the two claimed solutions, they look entirely different; I think you solved a different problem in Symbolab.

 

[Aminta_1900]

I meant, "If you are not required to solve for y, then this is valid."

Let's fill in a gap or two, and then solve for y, as you showed Symbolab did.

First, what you did (aiming at your book's answer, apparently) was to multiply by 4 and move C:
$$\frac{1}{4}x=\ln|\sin y|+C\\x=4\ln|\sin y|+4C\\x-4C=4\ln|\sin y|\\x+C_1=4\ln|\sin y|$$
At the last step, I set $C_1 = -4C$, which is just a different arbitrary constant.

To solve for y, we can now divide by 4 again, and exponentiate:
$$\frac{x+C_1}{4}=\ln|\sin y|\\e^{(x+C_1)/4}=e^{\ln|\sin y|}\\e^{(x+C_1)/4}=|\sin y|\\\sin y=\pm e^{(x+C_1)/4}$$
We could then use the inverse sine to solve. But Symbolab apparently prefers to use the inverse cosine, probably because it makes it a little easier to show all solutions. That's not something I would have thought of, and I don't at a glance see how they got $4x+1$; but maybe we can do that using the fact that $\cos^2(x)=1-\sin^2(x)$. Let's see:
$$\sin^2y=e^{2(x+C_1)/4}\\1-\sin^2y=1-e^{(x+C_1)/2}\\\cos^2y=1-e^{(x+C_1)/2}\\\cos y=\pm\left(1-e^{(x+C_1)/2}\right)^{1/2}$$That doesn't work.

Actually, when I graph the two claimed solutions, they look entirely different; I think you solved a different problem in Symbolab.

Thanks a lot for that, I'll certainly brush up on my logarithms in future!