I need help with homework:
Let $T'(x)=45-0.75\cdot T(x)$, where $T(0)=75$. Find $T(x)$.
Here is what I have tried:
Let $T(x)=y.$ Rewrite the equation
$$\frac{\mathrm dy}{\mathrm dx}=45-0.75y\implies \frac{\mathrm dy}{45-0.75y}=\mathrm dx$$
Integrate both sides
$$\int \frac{\mathrm dy}{45-0.75y}=\int \mathrm dx$$
Calculate LHS
$$-\frac43\ln(45-0.75y)+c=x$$
Now isolate $y$. Divide by "$-\frac43$"
$$\ln(45-0.75y)+c=-\frac34x$$
Raise both sides in "$e$"
$$45-0.75y+c=e^{-\frac34x}$$
Subtract $45$ and $c$, divide by "$-0.75$"
$$T(x)=y=\frac{e^{-3x/4}-45}{0.75}-c$$
$$T(0)=75\implies c=16.33$$
$$T(x)=\frac{e^{-3x/4}-45}{0.75}-16.33$$
I know that the true answer is $$T(x)=15e^{-0.75 x}+60$$
However I am unsure on how to get that result.
Best Answer
I think you missed an absolute value (note that $45-0.75\cdot 75<0$).
If we separate the variables, we have that $$\int\frac{dT}{45-0.75\cdot T}=\int 1dx$$ which gives $$\frac{\ln(|45-0.75\cdot T|)}{-0.75}=x+c.$$ By letting $T=75$ and $x=0$ we get $c=-\ln(45/4)/0.75$.
Can you take it from here?
P.S. As suggested by Aleksas Domarkas, you may also solve the linear ODE with constant coefficients $T′+0.75T=45$ whose general solution is $$T(x)=Ce^{-0.75 x}+60.$$ Now by using the initial value $T(0)=75$ to find $C=15$ and the solution is $$T(x)=15e^{-0.75 x}+60.$$