# 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?

$$\displaystyle\frac{{d^2y}}{{\left.{d}{x}\right.}^{{2}}}−{2}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{10}{y}={0}$$ 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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Bentley Leach

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 $$\displaystyle{y}={\left(\frac{{{4}{e}^{{-{{x}}}}}}{{3}}\right)}{\sin{{\left({3}{x}\right)}}}$$