Differential equation $a+bf(x)$ We have the DE $$\frac{\operatorname{d}}{\operatorname{d}x}f(x)=a+bf(x)$$
Solve differential equation by substituting $u(x)=a+bf(x)$ and solve for $u(x)$ knowing that $a$ and $b$ are constant. Hint "$e$".
I've looked at the answer stating $f(x)= Ce^{xb}-\frac{a}{b}$, however I do not understand how this is done.
Sorry for format, holiday without computer only mobile.
 A: We know that $f'(x) = a + bf(x)$. If we instead consider the function $u(x) = a + bf(x)$, then 
$$u'(x) = bf'(x) = b(a + bf(x)) = bu(x).$$
So we find that we want to solve the differential equation
$$u'(x) = bu(x),$$
which is a classical, separable differential equation, whose solution is
$$u(x) = Ce^{bx},$$
where $C$ is any generic constant. As $u(x) = a + bf(x)$, and we want $f(x)$, we rewrite this relation as $f(x) = \dfrac{u(x) - a}{b}$ to see that the overall answer is
$$f(x) = Ce^{bx} - \frac{a}{b},$$
(and you might note that I've kept $C$ as $C$, instead of $C/b$, because it's still a generic constant).
A: If you do the substitution $u(x)=a+bf(x)$, you have:
$$\frac{d}{dx}u(x)=b\frac{d}{dx}f(x)$$
Then the DE turns into:
$$\frac{1}{b}\frac{d}{dx}u(x)=u(x)$$
So,
$$\begin{array}{rcl}
\frac{d}{dx}u(x)&=&bu(x)\\
\Rightarrow \frac{du}{u}&=&bdx\qquad /\int\\
\ln(u)&=&bx+K
\end{array}$$
where $K$ is the integration constant.
Now, solving for $u(x)$, we have:
$$\begin{array}{rcl}
\ln(u)&=&bx+K\qquad/\,\mathrm{exp}()\\
\Rightarrow u(x)&=&\mathrm{e}^{bx+K}\\
&=&C\mathrm{e}^{bx}
\end{array}$$
where $C=\mathrm{e}^K$. Now, solve for $f(x)$:
$$\begin{array}{rcl}
u(x)&=&C\mathrm{e}^{bx}\\
a+bf(x)&=&C\mathrm{e}^{bx}\\
bf(x)&=&C\mathrm{e}^{bx}-a\\
f(x)&=&\frac{C}{b}\mathrm{e}^{bx}-\frac{a}{b}\\
\end{array}$$
A: We see that $y' = a + by$ and we set $u=a+by$. Note that $u' = by'$. Now we have $$\frac{u'}{b} = u \Rightarrow \frac{du}{dx} \frac{1}{b} = u \Rightarrow \frac{1}{u} du = b dx \Rightarrow \int \frac{1}{u} du = \int b dx \Rightarrow \ln u = bx + C' \\ \Rightarrow \ln (a + by) = bx + C' \Rightarrow a + by = e^{bx + C'} \Rightarrow y = \frac{e^{bx+C'}}{b} -\frac{a}{b} \Rightarrow y = Ce^{bx} - \frac{a}{b} $$
A: The first thing to note about the substitution ($u(x)=a+bf(x)$) is the $\frac{\mathrm{d}}{\mathrm{d}x}f(x) = \frac{1}{b}\frac{\mathrm{d}}{\mathrm{d}x}u(x)$. See why?
From there we have the DE:
$$\frac{\mathrm{d}}{\mathrm{d}x}u(x)=bu(x)$$
which has a solution $u(x)=C'e^{bx}$.
But remember that $$f(x) = \frac{u(x)-a}{b}=\frac{C'e^{bx}}{b}-\frac{a}{b}$$
By saying that $C=\frac{C'}{b}$ we have
$$f(x)=Ce^{bx}-\frac{a}{b}$$
A: For shorthand I'll use $f$ to represent $f(x)$ and $u$ to represent $u(x)$
So we are given: $\frac{df}{dx}=a+bf\tag{1}$
And you were given a hint to use: $u=a+bf\tag{2}$
Differentiating (2) gives: $\frac{du}{dx}=b\frac{df}{dx}\tag{3}$
Substituting (2) and (3) into (1) gives: $\frac{1}{b}\frac{du}{dx}=u\tag{4}$
Can you solve from here?
