Results 1 to 4 of 4

Thread: initial value,Evaluate the integral problems

  1. #1
    Junior Member
    Joined
    Sep 2005
    Posts
    62

    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:


    Answer: ______ + c

    and

    Perform the following integration:



    Answer: ______ + c
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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,
    $\displaystyle \boxed{ y= \frac{1}{5}e^{2x}+\frac{4}{5}e^{-3x} }$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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,
    $\displaystyle 2\int \cosh u du=2\sinh u+C=2\sinh \sqrt{x}+C$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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,
    $\displaystyle \frac{1}{4}\sec (4t-1)+C$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Initial Value Problems
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: Sep 15th 2009, 04:59 PM
  2. initial value problems
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jan 31st 2009, 10:39 PM
  3. Initial Value Problems
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 2nd 2008, 03:34 PM
  4. initial value problems
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 25th 2007, 06:44 PM
  5. Initial value problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 30th 2007, 10:58 PM

Search Tags


/mathhelpforum @mathhelpforum