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Hey guys I'm stuck with a Differential equation.

Can anyone help me finding the complete solution to $dy/dx + 2y = 3e^x$

Any help would be greatly appreciated! :)

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closed as off-topic by Did, Claude Leibovici, Watson, zz20s, user228113 Apr 27 '16 at 17:03

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    $\begingroup$ Do you know how to solve $y'+2y=0$? $\endgroup$ – almagest Apr 26 '16 at 19:32
  • $\begingroup$ No, sadly I dont. I just started math again after working for 7 years, so It all gone. $\endgroup$ – Simon Nielsen Apr 26 '16 at 19:38
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    $\begingroup$ Ok. The equation $y'+k y=0$, where $k$ is a constant, has general solution $y=Ae^{-kx}$, as you can easily check by differentiating. To deal with the term on the rhs you try $e^x$. Substituted into the lhs that gives $e^x+2e^x$ which is what you want. Hence the general solution is $y=e^x+Ae^{-2x}$. You find the value of $A$ by an initial condition which might specify the value of $y(0)$. $\endgroup$ – almagest Apr 26 '16 at 19:43
  • $\begingroup$ Ok. I go the most part. But where do you get the e^x? $\endgroup$ – Simon Nielsen Apr 26 '16 at 19:52
  • $\begingroup$ You just have to know that the exponential function $e^x$ has the property that differentiating it gives $e^x$. So if you differentiate $Ae^x$ you just get $Ae^x$. If you differentiate $Ae^{kx}$ you get $Ake^{kx}$. $e$ is a number around 2.71828. $\endgroup$ – almagest Apr 26 '16 at 20:00
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Solving differential equations of this kind consists of two parts:

  1. finding the general solution $y_c(x)$ of the homogeneous differential equation $$y_c' + 2y_c = 0$$

  2. finding any one solution $y_p(x)$ that satisfy $$y_p' + 2y_p = 3e^x$$


To solve $y_c(x)$, notice the property of exponential function $$\frac d{dx}Ce^{kx} = Cke^{kx}$$ So by trying $y_c(x) = Ce^{kx}$, $$y_c' + 2y_c = Cke^{kx}+2Ce^{kx} = C(k+2)e^{kx}$$ Regardless of the value of $C$, and since $e^{kx}\ne 0$, this becomes solving a polynomial equation $$k+2 = 0$$ And so the general solution for $y_c$ is $$y_c(x) = Ce^{-2x}$$


A rule of thumb function to set $y_p(x)$ with is $C_2e^x$, which is the right hand side $3e^x$ scaled by a factor. Then

$$y_p'+2y_p = C_2e^x + 2C_2e^x = 3C_2e^x$$

which gives $C_2 = 1$. This time, however, $C_2$ is not an arbitrary constant, because a scaled $y_p$ does not give the same right hand side $3e^x$.


The required general $y(x) = y_c(x) + y_p(x) = Ce^{-2x} + e^x$, because $$(y_c+y_p)' + 2(y_c+y_p) = y_c'+2y_c + y_p'+2y_p = 0 + 3e^x = 3e^x$$

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