# Solution verification on homework problem. Separable first order ODE IVP.

The answer is supposedly $$y^2 = 1 + \sqrt{x^2 - 16}$$ I don't know where I went wrong cause I know for a fact that my substitution of $$x = 4 \sec(\theta)$$ is correct. I know for a fact that after substitution the integral becomes $$\int \sec^2(\theta)= \tan(\theta)+C$$. I am pretty sure that my substitution of $$\tan(\theta)$$ is correct. Am I missing something?

$$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$$

$$\int 2y \space dy = \int \frac{4\sec(\theta)}{\sqrt{(4\sec(\theta))^2-16}}\space d(\theta)$$

$$y^2 = \int \frac{4\sec(\theta)}{\sqrt{(\sec^2(\theta)-1)16}}\sec(\theta) \ tan(\theta)$$

$$y^2 = \frac{4\sec(\theta)}{4\tan(\theta)}\sec(\theta)\tan(\theta)$$

$$y^2 = \int sec^2(\theta)$$

$$y^2 = \tan(\theta) + c$$

Using the reference triangle: Tangent is equal to $$\frac{\sqrt{x^2-16}}{4}$$

$$y^2= \frac{\sqrt{x^2-16}}{4} + C$$

$$4 = \frac{3}{4}+ C$$

$$C = \frac{15}{4}$$

$$y^2= \frac{\sqrt{x^2-16}}{4} + \frac{15}{4}$$

You have to use another $$4$$ .
$$x = 4\sec\theta \implies dx = \color{red}4\sec\theta\tan\theta d\theta$$

So, you'll have $$y^2 = \sqrt{x^2-16} +c$$

$$2^2 = \sqrt{9} +c \implies c = 4-3= 1$$

Thus,

$$y^2 =\sqrt{x^2+16} +1$$

In integrating the RHS of in$$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$$

it is advised to let $$u=x^2-16$$ and you have $$du=2xdx$$

Then you do not have to worry about trig substitution at all.