Question # Solve the for the general solution of the differential equation. (D^{2}+2D-1)y=0

Equations
ANSWERED Solve the for the general solution of the differential equation.
$$\displaystyle{\left({D}^{{{2}}}+{2}{D}-{1}\right)}{y}={0}$$ 2021-04-11
Step 1
We have to solve the differential equation:
$$\displaystyle{\left({D}^{{{2}}}+{2}{D}-{1}\right)}{y}={0}$$
The given equation is in symbolic form
where,
$$\displaystyle{D}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle{D}^{{{2}}}={\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}$$
Its auxiliary equation would be:
$$\displaystyle{D}^{{{2}}}+{2}{D}-{1}={0}$$
Solving using the formula of quadratic equation,
here,
a=1
b=2
c=-1
Therefore,
$$\displaystyle{D}={\frac{{-{b}\pm\sqrt{{{b}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}$$
$$\displaystyle={\frac{{-{2}\pm\sqrt{{{2}^{{{2}}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{{2}\times{1}}}}$$
$$\displaystyle={\frac{{-{2}\pm\sqrt{{{4}+{4}}}}}{{{2}}}}$$
$$\displaystyle={\frac{{-{2}\pm\sqrt{{{8}}}}}{{{2}}}}$$
$$\displaystyle={\frac{{-{2}\pm{2}\sqrt{{{2}}}}}{{{2}}}}$$
$$\displaystyle={\frac{{{2}{\left({1}+\pm\sqrt{{{2}}}\right)}}}{{{2}}}}$$
$$\displaystyle={1}\pm\sqrt{{{2}}}$$
$$\displaystyle={1}+\sqrt{{{2}}},{1}-\sqrt{{{2}}}$$
Step 2
Hence, there are two solution of auxiliary equation
$$\displaystyle{a}={1}+\sqrt{{{2}}}$$
$$\displaystyle{b}={1}-\sqrt{{{2}}}$$
We know solution for this type of differential equation is given as:
$$\displaystyle{y}={c}_{{{1}}}{e}^{{{a}{x}}}+{c}_{{{2}}}{e}^{{{b}{x}}}$$
where, $$\displaystyle{c}_{{{1}}}{\quad\text{and}\quad}{c}_{{{2}}}$$ are arbitrary constants.
Putting values of a and b,
$$\displaystyle{y}={c}_{{{1}}}{e}^{{{a}{x}}}+{c}_{{{2}}}{e}^{{{b}{x}}}$$
$$\displaystyle{y}={c}_{{{1}}}{e}^{{{\left({1}+\sqrt{{{2}}}\right)}{x}}}+{c}_{{{2}}}{e}^{{{\left({1}-\sqrt{{{2}}}\right)}{x}}}$$
Hence, solution of given differential equation is $$\displaystyle{y}={c}_{{{1}}}{e}^{{{\left({1}+\sqrt{{{2}}}\right)}{x}}}+{c}_{{{2}}}{e}^{{{\left({1}-\sqrt{{{2}}}\right)}{x}}}$$.