d^2y/dx^2−2dy/dx+10y=0 where x=0.y=0 and dy/dx=4?x=0.y=0 and dy/dx=4?

Daniaal Sanchez 2021-08-12 Answered

\(\displaystyle\frac{{d^2y}}{{\left.{d}{x}\right.}^{{2}}}−{2}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{10}{y}={0}\) where x=0.y=0 and \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={4}?{x}={0}.{y}={0}\) and \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={4}\)?

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Expert Answer

Bentley Leach
Answered 2021-08-13 Author has 16948 answers

The given differential is \(\displaystyle\frac{{{d}^{{2}}y}}{{\left.{d}{x}\right.}^{{2}}}-{2}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{10}{y}={0}\) where x=0.y=0 and \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={4}\).
The auxilary equation is given by \(\displaystyle{m}^{{2}}-{2}{m}+{10}={0}\)
\(\displaystyle\to{m}=-{2}+{\left(-\sqrt{{{4}-{40}}}\right)}\to{m}=-{2}\pm{2}{\left(\sqrt{{{1}-{10}}}\right)}\to{m}=-{1}\pm\sqrt{{{1}-{10}}}\to{m}=-{1}\pm\sqrt{{9}}\to{m}=-{1}\pm{3}{i}\)
Therefore the general olution is given by \(\displaystyle{y}{\left({x}\right)}={e}^{{-{{x}}}}{\left({c}_{1}{\cos{{\left({3}{x}\right)}}}+{c}_{2}{\sin{{\left({3}{x}\right)}}}\right)}\)
When \(x=0,y=0\) therefore \(c_1=0\). Therefore \(\displaystyle{y}={c}_{2}{e}^{ -{{x}}}{\left( \sin{{\left({3}{x}\right)}}\right)}\to\frac{{\left.{d}{y}\right.}}{{x}}={c}_{2}{e}^{ -{{x}}}{\left(- \sin{{\left({3}{x}\right)}}+{3} \cos{{\left({3}{x}\right)}}\right)}\)
Again when \(\displaystyle{x}={0},\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={4}\), this implies that \(4=3e^2\)
Therefore the solution is \(\displaystyle{y}={\left(\frac{{{4}{e}^{{-{{x}}}}}}{{3}}\right)}{\sin{{\left({3}{x}\right)}}}\)

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