# Does this look correct? (DE)

## Homework Statement

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

## 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.

Last edited:

SteamKing
Staff Emeritus
Homework Helper
You can always differentiate your y expression and see if it satisfies the original DE.

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
$$\frac{1}{\sqrt{1- \frac{1}{D^2}\frac{1}{x^2+ 1}}}\frac{x}{D\sqrt{x^2+ 1}}= \frac{x}{\sqrt{D^2(x^2+ 1)- 1}}$$