• Support PF! Buy your school textbooks, materials and every day products Here!

Does this look correct? (DE)

  • Thread starter iRaid
  • Start date
  • #1
559
8

Homework Statement


[tex](x^{2}+1)(tan y)(y')=x[/tex]


Homework Equations





The Attempt at a Solution


[tex](x^{2}+1)(tan y)(\frac{dy}{dx})=x[/tex]
[tex](tan y)dy=\frac{x}{x^{2}+1}dx[/tex]
[tex]\int tanydy=\int \frac{x}{x^2+1}dx \\ -ln|cos y|=\frac{ln|x^{2}+1|}{2}+C[/tex]
[tex]e^{-ln(cosy)}=e^{\frac{ln(x^{2}+1)}{2}+C}[/tex]
[tex]\frac{1}{cosy}=e^{C}(\sqrt{x^2+1})[/tex]
[tex]cosy=\frac{1}{D\sqrt{x^{2}+1}} \implies y=cos^{-1}\bigg(\frac{1}{D\sqrt{x^2+1}}\bigg)[/tex]
D represents e^C just fyi.

Wondering if this is correct and if my work makes sense.
 
Last edited:

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
You can always differentiate your y expression and see if it satisfies the original DE.
 
  • #3
559
8
I don't even know how to differentiate that LOL
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,777
911
Strange! You are solving differential equations but don't know how to differentiate? The derivative of arc-cosine(x) is [itex]1/\sqrt{1- x^2}[/itex] and the derivative of [itex](x^2+ 1)^{1/2}[/itex] is [itex](1/2)(x^2+ 1)^{-1/2}(2x)= x(x^2+ 1)^{-1/2}= x/\sqrt{x^2+ 1}[/itex].

Putting those together with the chain rule, the derivative of [itex]cos^{-1}(1/D\sqrt{x^2+ 1})[/itex] is
[tex]\frac{1}{\sqrt{1- \frac{1}{D^2}\frac{1}{x^2+ 1}}}\frac{x}{D\sqrt{x^2+ 1}}= \frac{x}{\sqrt{D^2(x^2+ 1)- 1}}[/tex]
 

Related Threads for: Does this look correct? (DE)

  • Last Post
Replies
6
Views
551
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
3
Views
961
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
853
  • Last Post
Replies
3
Views
796
Top