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Can someone help me to solve this? $$V\left(\frac{\mathrm dc}{\mathrm dx}\right)=w-Qc-KcV$$

$w,Q,K,V$ are constants

I think it is linear but i can't get the correct answer which is $$c(x)=[w/(Q+KV)](1-e^{-(Q/V+K)x})$$

Thanks in advance

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    Did I get the exponent to the exponential function right? – 2017-02-05
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    @mathreadler yes the exponent is right but the derivative is not partial – 2017-02-05
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    @van What you have done?, should add in post. – 2017-02-05
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    @van : Why do you make your life harder in using a lot of non-mathematical constants ? Is it not simpler for you to solve the equation $$\frac{dy}{dx}=ay+b$$ where $y(x)=c(x)\quad;\quad a=w/V \quad;\quad b=-(Q+KV)/V$. – 2017-02-05
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    @J Jacquelin according to your replacements form should be $$\frac{\mathrm dy}{\mathrm dx}=by+a$$ – 2017-02-05
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    Since it is indeed linear, you can solve this using an [Integrating Factor](http://mathworld.wolfram.com/IntegratingFactor.html) or using [Variation of Parameters](https://en.m.wikipedia.org/wiki/Variation_of_parameters). – 2017-02-05

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Your ODE is equal to

$$\frac{dc}{dx}=\frac{-Q-KV}{V}c+\frac{w}{V}$$

Set $$ f(c)=\frac{-Q-KV}{V}c+\frac{w}{V} $$ then $ f(\frac{w}{Q+KV})=0 $ hence $c_s(x)=\frac{w}{Q+KV}$ is a special solution of the ODE. Now consider $$\frac{dc}{dx}=\frac{-Q-KV}{V}c = Ac \ , \ \ A=\frac{-Q-KV}{V}$$ Here the general solution is $c_g(x)=Ae^{Ax}$. Thus

$$c(x)=c_s(x)+c_g(x)$$ is the solution, which is exactly the solution you wrote in the last line.