# initial value,Evaluate the integral problems

• Feb 4th 2007, 02:45 PM
killasnake
initial value,Evaluate the integral problems
Hi I need help with these problems, I just started calc 2 and I have no idea how to do these. :confused:

Solve the initial value problem:
http://img.photobucket.com/albums/v629/killasnake/1.gif

and y=1 when x=0

y(x)= ___________

and

Evaluate the integral:
http://img.photobucket.com/albums/v6...241696img1.gif

and

Perform the following integration:

http://img.photobucket.com/albums/v629/killasnake/0.gif

• Feb 4th 2007, 03:52 PM
ThePerfectHacker
You have,
$\displaystyle y'+3y=e^{2x}$
An integrating factor is,
$\displaystyle y=\exp \left( \int 3 dx \right)=e^{3x}$.
Thus, the general solution is,
$\displaystyle y=e^{-3x}\int e^{3x}e^{2x} dx$
$\displaystyle y=e^{-3x}\int e^{5x} dx$
$\displaystyle y=e^{-3x} \left( \frac{1}{5}e^{5x}+C \right)$
$\displaystyle y=\frac{1}{5}e^{2x}+Ce^{-3x}$
We also know that, $\displaystyle y(0)=1$.
Thus,
$\displaystyle 1=\frac{1}{5}(1)+C(1)$
$\displaystyle C=\frac{4}{5}$.
Thus,
$\displaystyle \boxed{ y= \frac{1}{5}e^{2x}+\frac{4}{5}e^{-3x} }$
• Feb 4th 2007, 03:54 PM
ThePerfectHacker
You have,
$\displaystyle \int \frac{\cosh \sqrt{x}}{\sqrt{x}}dx=2\int \frac{\cosh \sqrt{x}}{2\sqrt{x}} dx$
Use substitution $\displaystyle u=\sqrt{x}$.
Thus,
$\displaystyle 2\int \cosh u du=2\sinh u+C=2\sinh \sqrt{x}+C$
• Feb 4th 2007, 03:58 PM
ThePerfectHacker
First in,
$\displaystyle \int \frac{\sin (4t-1)}{1-\sin^2 (4t-1)} dt$
Use the linear substitution $\displaystyle x=4t-1$ to get,
$\displaystyle \frac{1}{4} \int \frac{\sin x}{1-\sin^2 x} dx$
Pythagorean identity,
$\displaystyle \frac{1}{4} \int \frac{\sin x}{\cos^2 x} dx$
Let $\displaystyle u=\cos x$ then $\displaystyle u'=-\sin x$.
Thus,
$\displaystyle -\frac{1}{4} \int \frac{1}{u^2} du$
Thus,
$\displaystyle \frac{1}{4u}+C=\frac{1}{4\cos x}+C=\frac{1}{4}\sec x+C$
Substitute again,
$\displaystyle \frac{1}{4}\sec (4t-1)+C$