Ytukyg

I need help with homework:

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

Here is what I have tried:

Let $T(x)=y.$ Rewrite the equation
$$frac{mathrm dy}{mathrm dx}=45-0.75yimplies frac{mathrm dy}{45-0.75y}=mathrm dx$$
Integrate both sides
$$int frac{mathrm dy}{45-0.75y}=int mathrm dx$$
Calculate LHS
$$-frac43ln(45-0.75y)+c=x$$
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 "$-0.75$"
$$T(x)=y=frac{e^{-3x/4}-45}{0.75}-c$$
$$T(0)=75implies c=16.33$$
$$T(x)=frac{e^{-3x/4}-45}{0.75}-16.33$$

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 $1/(45-0,75x)$

I think you missed an absolute value (note that $45-0.75cdot 75<0$).

If we separate the variables, we have that
$$intfrac{dT}{45-0.75cdot T}=int 1dx$$
which gives
$$frac{ln(|45-0.75cdot T|)}{-0.75}=x+c.$$
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
$$T(x)=15e^{-0.75 x}+60.$$

Your first grave error is that you used a wrong exponentiation rule. The exponential of $$ln(u)+c$$ is $$e^ccdot u$$ 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.