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?

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- Sep 8th 2009, 08:48 PMgenlovesmusic09d2y/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?

- Sep 8th 2009, 08:51 PMpickslides
$\displaystyle y = \int \tan(x)+1 ~dx= \int \frac{\sin(x)}{\cos(x)}+1 ~dx$

Can you take it from here? - Sep 8th 2009, 09:01 PMgenlovesmusic09
i know that the antiderivative of sinx/cosx is -ln cosx +c but wher does the +1 come in?

- Sep 9th 2009, 02:03 PMpickslides
$\displaystyle 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

$\displaystyle

\int 1 ~dx = x+c

$

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

$

Now use initial condition of $\displaystyle y(0)=0 $ to solve for c. - Sep 9th 2009, 03:00 PMgenlovesmusic09
thanks! i forgot you can separate!