What is the solution for (dq)/(20-q)= 1/16 dT ?

The solution for (dq)/(20-q)= 1/16 dT is T(q)=c_{1}-16d(-20+q+20ln(20-q))

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(dq)/(20-q)= 1/16 dT

\frac{d q}{20 - q} = \frac{1}{16} d T

T (q) = c_{1} - 16 d (- 20 + q + 20 ln (20 - q))

Solution steps

\frac{d q}{20 - q} = \frac{1}{16} d T (q)

Solve linear ODE:

T (q) = c_{1} - 16 d (- 20 + q + 20 ln (20 - q))

T (q) = c_{1} - 16 d (- 20 + q + 20 ln (20 - q))

y^{^{''}} - y^{^{'}} + 4 y = 4, e^{2 x} y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{2 x} + 1

4 y^{^{''}} - y = e^{\frac{x}{2}} x

\frac{d ^{2} y}{d x ^{2}} + y = 3 sin (2 x) + x cos (2 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{4 x}

What is the solution for (dq)/(20-q)= 1/16 dT ?
The solution for (dq)/(20-q)= 1/16 dT is T(q)=c_{1}-16d(-20+q+20ln(20-q))