# Thread: initial value,Evaluate the integral problems

1. ## 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.

Solve the initial value problem:

and y=1 when x=0

y(x)= ___________

and

Evaluate the integral:

and

Perform the following integration:

2. 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,
$\boxed{ y= \frac{1}{5}e^{2x}+\frac{4}{5}e^{-3x} }$

3. 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,
$2\int \cosh u du=2\sinh u+C=2\sinh \sqrt{x}+C$

4. 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,
$\frac{1}{4}\sec (4t-1)+C$