How Do You Solve This Trigonometric Differential Equation?

[iRaid]

Homework Statement

(x^{2}+1)(tan y)(y')=x

Homework Equations

The Attempt at a Solution

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

Wondering if this is correct and if my work makes sense.

 

[SteamKing]

You can always differentiate your y expression and see if it satisfies the original DE.

 

[iRaid]

I don't even know how to differentiate that LOL

 

[HallsofIvy]

Strange! You are solving differential equations but don't know how to differentiate? The derivative of arc-cosine(x) is 1/\sqrt{1- x^2} and the derivative of (x^2+ 1)^{1/2} is (1/2)(x^2+ 1)^{-1/2}(2x)= x(x^2+ 1)^{-1/2}= x/\sqrt{x^2+ 1}.

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