initial value,Evaluate the integral problems

**killasnake** · February 4th 2007, 02:45 PM

Hi I need help with these problems, I just started calc 2 and I have no idea how to do these.

Solve the initial value problem:

and y=1 when x=0

y(x)= ___________

and

Evaluate the integral:

Answer: ______ + c

and

Perform the following integration:

Answer: ______ + c

**ThePerfectHacker** · February 4th 2007, 03:52 PM

You have,
$y'+3y=e^{2x}$
An integrating factor is,
$y=\exp \left( \int 3 dx \right)=e^{3x}$ .
Thus, the general solution is,
$y=e^{-3x}\int e^{3x}e^{2x} dx$
$y=e^{-3x}\int e^{5x} dx$
$y=e^{-3x} \left( \frac{1}{5}e^{5x}+C \right)$
$y=\frac{1}{5}e^{2x}+Ce^{-3x}$
We also know that, $y(0)=1$ .
Thus,
$1=\frac{1}{5}(1)+C(1)$
$C=\frac{4}{5}$ .
Thus,
$\boxed{ y= \frac{1}{5}e^{2x}+\frac{4}{5}e^{-3x} }$

**ThePerfectHacker** · February 4th 2007, 03:54 PM

You have,
$\int \frac{\cosh \sqrt{x}}{\sqrt{x}}dx=2\int \frac{\cosh \sqrt{x}}{2\sqrt{x}} dx$
Use substitution $u=\sqrt{x}$ .
Thus,
$2\int \cosh u du=2\sinh u+C=2\sinh \sqrt{x}+C$

**ThePerfectHacker** · February 4th 2007, 03:58 PM

First in,
$\int \frac{\sin (4t-1)}{1-\sin^2 (4t-1)} dt$
Use the linear substitution $x=4t-1$ to get,
$\frac{1}{4} \int \frac{\sin x}{1-\sin^2 x} dx$
Pythagorean identity,
$\frac{1}{4} \int \frac{\sin x}{\cos^2 x} dx$
Let $u=\cos x$ then $u'=-\sin x$ .
Thus,
$-\frac{1}{4} \int \frac{1}{u^2} du$
Thus,
$\frac{1}{4u}+C=\frac{1}{4\cos x}+C=\frac{1}{4}\sec x+C$
Substitute again,
$\frac{1}{4}\sec (4t-1)+C$

Thread: initial value,Evaluate the integral problems

LinkBack

Thread Tools

Display

initial value,Evaluate the integral problems

Similar Math Help Forum Discussions

Initial Value Problems

initial value problems

Initial Value Problems

initial value problems

Initial value problems

Search Tags