Differential Equation Exercise
Consider the first-order differential equation with initial condition:
$$ \begin{cases} y' = y - 1 \\ \\ y(0) = 7 \end{cases} $$
This is a Cauchy problem, since the value $y(0) = 7$ is prescribed.
It is also a linear differential equation of the form $y' + a(x)y = b(x)$, here with $a(x) = -1$ and $b(x) = -1$:
$$ y' - y = -1 $$
We solve it using the method of variation of constants:
$$ y = e^{-\int a(x)\, dx}\,\Bigg(\int b(x)\, e^{\int a(x)\, dx}\, dx + c \Bigg). $$
Substituting $a(x) = -1$ and $b(x) = -1$ gives
$$ y = e^{-\int -1\, dx}\,\Bigg(\int -1 \cdot e^{\int -1\, dx}\, dx + c \Bigg). $$
Since $\int -1\, dx = -x$, this becomes
$$ y = e^x \,\Big( - \int e^{-x}\, dx + c \Big). $$
The integral $\int e^{-x}\, dx = -e^{-x}$, so
$$ y = e^x \,\big( -(-e^{-x}) + c \big) $$
$$ y = e^x \,( e^{-x} + c ) $$
$$ y = 1 + c e^x. $$
Thus, the general solution is
$$ y = c e^x + 1. $$
Applying the initial condition $y(0) = 7$:
$$ 7 = c e^0 + 1 \quad \Rightarrow \quad c = 6. $$
Hence, the particular solution of the Cauchy problem is
$$ y = 6e^x + 1. $$
Verification. The solution satisfies $y(0) = 7$.
Substituting $y = 6e^x + 1$ into $y' = y - 1$ gives $$ y' = y - 1, \qquad D(6e^x + 1) = (6e^x + 1) - 1, $$ $$ 6e^x = 6e^x, $$ confirming the identity.
Alternative approach
The equation can also be solved by separation of variables:
$$ \frac{dy}{dx} = y - 1 $$
$$ \frac{1}{y-1}\, dy = dx. $$
Integrating both sides,
$$ \int \frac{1}{y-1}\, dy = \int dx $$
gives
$$ \log|y-1| = x + c. $$
Exponentiating,
$$ y - 1 = \pm e^{x+c}. $$
Using the initial condition $y(0) = 7$:
$$ 6 = \pm e^c. $$
Since the right-hand side must be positive, we select the plus sign:
$$ y - 1 = e^{x+c}. $$
Note. The $\pm$ can be absorbed into the constant: $$ y - 1 = \pm e^{x+c} = (\pm e^c)\, e^x = c e^x. $$ Thus, the solution simplifies to $$ y - 1 = c e^x. $$
Hence, the general solution is
$$ y = c e^x + 1, $$
and with $c = 6$ we recover the same particular solution:
$$ y = 6e^x + 1. $$
