Undergrad Heat Equation Problem: Solving c + 3d = 0

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The discussion centers on solving the equation -c + 3d = 0 in the context of the heat equation. The equation arises from applying Fourier's law of heat conduction, which relates heat flow to temperature gradients. Perfect thermal contact at x=1 implies equal heat flow from both sides, leading to the equation derived from the constants in the derivatives. The participants clarify that substituting the given conductivities and temperature solutions will resolve the confusion. Understanding these relationships is crucial for solving heat equation problems effectively.
FAS1998
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I’ve attached an image of a solved problem related to the heat equation.

Can somebody explain the -c + 3d = 0 comes from? I’m having trouble following the work shown.
 

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Hard to say since your image is way too small to read
 
Perfect thermal contact occurs at x=1, so the flow coming from the right is equal to the flow coming from the left. Now recall that the flow is given by ##\phi=-K_0\frac{\partial u}{\partial x}##, that is Fourier's law. The derivatives on both sides are constants as it's shown on your work so we're just left with: $$K_{1}\dfrac{\partial u_1}{\partial x} = K_{2}\dfrac{\partial u_2}{\partial x} \\ -c = -3d \\ -c+3d=0$$ I took the liberty to introduce some notation since it's a little confusing in your solution, but essentially all you have to do is substitute the conductivities given in the exercise and the solutions you found for u.
 
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