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$$ \frac{dy}{dx}= K \left(\frac{1}{\frac{1}{A}+Kx}\right)^2 -Cy $$

$K$, $A$ and $C$ are constants.

The problem is that the variables $y$ and $x$ are not separable. I want to express y explicitly in terms of $x$ and the constants. I would greatly appreciate any guidance. Thank you.

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    What is * ? in your formula ? – 2017-01-13
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    its the multiplication symbol (ie times). I just removed it to avoid confusion. – 2017-01-13
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    @Sp500 Do you want the equation to read $$\frac{\mathrm{d}y}{\mathrm{d}x}=K\left(\frac{1}{\frac{1}{A}+Kx}\right)^2-Cy $$? – 2017-01-13
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    Yes , exactly that. – 2017-01-13

1 Answers 1

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You wish to solve the ODE

$$y' + Cy = \frac{K}{(m+Kx)^2}$$

where $m:=\frac{1}{A}$.

This is a linear, nonhomogeneous equation. We first solve the associated homogeneous equation

$$y' + Cy=0$$

which has the solution

$$y(x) = Be^{-Cx} \tag1$$

for some arbitrary constant $B$.

To solve the nonhomogeneous problem, assume that this constant is a function of $x$ instead, that is $B=B(x)$. Plugging this into our differential equation, we have

$$B'(x)e^{-Cx}-B(x)Ce^{-Cx}+B(x)Ce^{-Cx}=\frac{K}{(m+Kx)^2}\implies B'(x)=\frac{Ke^{Cx}}{(m+Kx)^2}$$

Thus we have

$$B(x) = K\int\frac{e^{cx}}{(m+Kx)^2}dx$$

This is not an elementary integral - it is given in terms of Exponential Integrals.

Multiplying what we have for $B(x)$ with $e^{-Cx}$ gives us a particular solution to our ODE. Adding the general solution $(1)$ and this particular solution gives you the complete solution of your differential equation.

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    Thank you so much. So the exponential term cancels out after the multiplication? – 2017-01-13
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    If you mean the cancellation of $-B(x)Ce^{-Cx}$ and $+B(x)Ce^{-Cx}$, then yes. That always happens - that's why the method of variation of constants works so well - the differential equation that you get for $B(x)$ can always been integrated directly (though, as you have seen, not always be evaluated elementarily). – 2017-01-13
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    Wow, it works. I am genuinely amazed. This method is beautiful! Thank you very much sir. – 2017-01-13
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    You're welcome. Don't forget to accept the answer if you feel satisfied with it. – 2017-01-13