# Math Help - d2y/dx2?

1. ## d2y/dx2?

d2y/dx2=sec 2x, y(0)=0 and y'(0)=1

so the first deritive is tanx+1
because tan0+c=1
c=1

for the orinal equation does it go back to sec 2x+c?

2. $y = \int \tan(x)+1 ~dx= \int \frac{\sin(x)}{\cos(x)}+1 ~dx$

Can you take it from here?

3. i know that the antiderivative of sinx/cosx is -ln cosx +c but wher does the +1 come in?

4. Originally Posted by genlovesmusic09

so the first deritive is tanx+1
$y = \int \tan(x)+1 ~dx= \int \frac{\sin(x)}{\cos(x)}+1 ~dx= \int \frac{\sin(x)}{\cos(x)} ~dx+ \int 1 ~dx
$

with

$
\int 1 ~dx = x+c
$

$y = \int \frac{\sin(x)}{\cos(x)} ~dx+ \int 1 ~dx = -\ln(-\cos(x)) +x+c
$

Now use initial condition of $y(0)=0$ to solve for c.

5. thanks! i forgot you can separate!