Solve the following differential equations. y" - 4y = 0

David Lewis 2021-12-31 Answered
Solve the following differential equations.
\(\displaystyle{y}\text{}-{4}{y}={0}\)

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Expert Answer

Linda Birchfield
Answered 2022-01-01 Author has 1383 answers
Given: \(\displaystyle{y}\text{}{4}{y}={0}\)
\(\displaystyle\rightarrow{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{4}{y}={0}\)
Let \(\displaystyle{\frac{{{d}^{{{2}}}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={0}^{{{2}}}=\) differential operator
\(\displaystyle{D}^{{{2}}}{y}-{4}{y}={0}\)
\(\displaystyle{\left({D}^{{{2}}}-{4}\right)}{y}={0}\)
For complementary sol put \(\displaystyle{D}^{{{2}}}-{4}={0}\)
\(\displaystyle{D}^{{{2}}}-{4}={0}\)
\(\displaystyle{D}^{{{2}}}={4}\)
\(\displaystyle{D}=\pm{2}\)
\(\displaystyle{m}_{{{1}}}={2}\ \text{and}\ {m}_{{{2}}}=-{2}\)
\(\displaystyle{y}={c}_{{{1}}}{e}^{{{m}{x}}}+{c}_{{{2}}}{e}^{{{m}{2}{x}}}\)
\(\displaystyle{c}_{{{1}}}\ \text{and}\ {c}_{{{2}}}=\text{constants}\)
Answer: \(\displaystyle{y}={c}_{{{1}}}{e}^{{{2}{x}}}+{c}_{{{2}}}{e}^{{-{2}{x}}}\)
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Chanell Sanborn
Answered 2022-01-02 Author has 4613 answers
\(\displaystyle{m}^{{{2}}}-{4}={0},{m}{1}={2},{m}{2}=-{2}\).
\(\displaystyle{y}={C}{1}\cdot{e}^{{{2}{x}}}+{C}{2}\cdot{e}^{{-{2}{x}}},{\quad\text{or}\quad}\ {y}={C}{1}\cdot{c}{h}{\left({2}{x}\right)}+{C}{2}\cdot{s}{h}{\left({2}{x}\right)}\)
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