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Math Help - initial value,Evaluate the integral problems

  1. #1
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    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
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  2. #2
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    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} }
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  3. #3
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    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
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  4. #4
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    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
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