Solve $T'(x)=45-0.75\cdot T(x)$, where $T(0)=75$. 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.

This is the integral calculator for $1/(45-0,75x)$
 A: 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
$$T(x)=15e^{-0.75 x}+60.$$
A: Your first grave error is that you used a wrong exponentiation rule. The exponential of $$\ln(u)+c$$ is $$e^c\cdot u$$ or $$Cu~~\text{ with  }~~C=\pm e^c,$$ not $u+c$ or $u+e^c$ or whatever. 
Changing the constant on the fly by incorporating other constant terms, which you do in practically every transformation while isolating $y$, is usually OK. But the number of changes should be kept to a minimum to avoid hard-to-reconstruct errors.
