Results 1 to 3 of 3

Math Help - y' + 3y = sin(x)

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    22

    y' + 3y = sin(x)

    Hi, can someone please check my work. I only have one chance to submit my answer online and I wanna be 100% sure.

    \frac{dx}{dy} + 3y = sin(x) , y(0) = 4
    e^3 dx  = e^{3x}
    e^{3x} \frac{dx}{dy} + e^{3x}y = e^{3x}sin(x) dx
    \int\frac{d}{dx} e^{3x}y = \int e^{3x}sin(x)
    e^{3x}y = \frac{e^{3x}3sin(x)-cos(x)}{10} + c
    y = \frac{3 sin(x) - cos(x)}{10}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,133
    Thanks
    810

    Re: y' + 3y = sin(x)

    Quote Originally Posted by JC05 View Post
    Hi, can someone please check my work. I only have one chance to submit my answer online and I wanna be 100% sure.

    \frac{dx}{dy} + 3y = sin(x) , y(0) = 4
    e^3 dx  = e^{3x}
    e^{3x} \frac{dx}{dy} + e^{3x}y = e^{3x}sin(x) dx
    \int\frac{d}{dx} e^{3x}y = \int e^{3x}sin(x)
    e^{3x}y = \frac{e^{3x}3sin(x)-cos(x)}{10} + c
    y = \frac{3 sin(x) - cos(x)}{10}
    this is a mess

    are you solving for an integrating factor? (yes) then make it clear.

    $$\mu(x)=e^{\int 3dx}=e^{3x}$$

    then

    $$\mu(x)\frac{dy}{dx}+\mu(x)y = \mu(x)\sin(x) \Rightarrow $$

    $$e^{3x}\frac{dy}{dx}+e^{3x}y=e^{3x}\sin(x)$$

    $$\frac{d}{dx}\left(e^{3x}y\right)=e^{3x}\sin(x)$$

    etc.

    Then, you properly introduced the integration constant C, but then lost it.

    You need to include that and solve for it given your initial condition.

    Rework it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    22

    Re: y' + 3y = sin(x)

    Quote Originally Posted by romsek View Post
    this is a mess

    are you solving for an integrating factor? (yes) then make it clear.

    $$\mu(x)=e^{\int 3dx}=e^{3x}$$

    then

    $$\mu(x)\frac{dy}{dx}+\mu(x)y = \mu(x)\sin(x) \Rightarrow $$

    $$e^{3x}\frac{dy}{dx}+e^{3x}y=e^{3x}\sin(x)$$

    $$\frac{d}{dx}\left(e^{3x}y\right)=e^{3x}\sin(x)$$

    etc.

    Then, you properly introduced the integration constant C, but then lost it.

    You need to include that and solve for it given your initial condition.

    Rework it.

    Ya, my bad...

    I figured it out.....

    \int\frac{d}{dx} e^{3x}y = \int e^{3x}sin(x)

    e^{3x} = \frac{e^{3x}(3sin(x) - cos(x)}{10} + C

    y = \frac{e^{3x}(3sin(x) - cos(x)}{10} + e^{-3x}C
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum