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?

Joni Kenny 2021-06-25 Answered

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

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Expert Answer

Gennenzip
Answered 2021-06-26 Author has 11659 answers

The given differential is \(\displaystyle{d}^{{2}}\frac{{y}}{{\left.{d}{x}\right.}^{{2}}}-{2}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{10}{y}={0}\) where \(x=0;y=0\) and \(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={4}\).
The auxilary equation is given by \(\displaystyle{m}^{{2}}-{2}{m}+{10}={0}\)
\(\rightarrow m=-2+(-\sqrt{4-40}) \rightarrow m=-2\pm2(\sqrt{1-10}) \rightarrow m=-1\pm \sqrt{1-10} \rightarrow m=-1 \pm \sqrt{9} \rightarrow m=-1\pm3i\)
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 \(y=c_2e^{-x}(\sin(3x)) \rightarrow dy/x=c_2e^{-x}(-\sin(3x)+3\cos(3x))\)
Again when \(\displaystyle{x}={0},\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={4}\), this implies that \(4=3e2\)
Therefore the solution is \(\displaystyle{y}={\left(\frac{{{4}{e}^{{-{{x}}}}}}{{3}}\right)}{\sin{{\left({3}{x}\right)}}}\)

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