What is the steady state solution for the heat conduction equation

Equipped with the following boundary conditions:
U(0, t) - (0, t) = 5
U(40, t) + (40, t) = 4
A) $ v(x)=\frac{1}{40} x+201 $
B) $ v(x)=\frac{1}{40} x+\frac{201}{40} $
C) $ v(x)=-\frac{1}{40} x+199 $
D) $ v(x)=-\frac{1}{40} x+\frac{199}{40} $

Question

What is the steady state solution for the heat conduction equation

Equipped with the following boundary conditions:
U(0, t) - (0, t) = 5
U(40, t) + (40, t) = 4
A) $ v(x)=\frac{1}{40} x+201 $ 
B) $ v(x)=\frac{1}{40} x+\frac{201}{40} $ 
C) $ v(x)=-\frac{1}{40} x+199 $ 
D) $ v(x)=-\frac{1}{40} x+\frac{199}{40} $

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The Answer of What is the steady state solution for the heat conduction equation

Equipped with the following boundary conditions:
U(0, t) - (0, t) = 5
U(40, t) + (40, t) = 4
A) $ v(x)=\frac{1}{40} x+201 $ 
B) $ v(x)=\frac{1}{40} x+\frac{201}{40} $ 
C) $ v(x)=-\frac{1}{40} x+199 $ 
D) $ v(x)=-\frac{1}{40} x+\frac{199}{40} $

What Is the Steady State Solution for the Heat Conduction v(x)=140x+201 v(x)=\frac{1}{40} x+201 v(x)=401​x+201

What Is the Steady State Solution for the Heat Conduction $v(x)=\frac{1}{40} x+201$