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

d2ydx22dydx+10y=0 where x=0;y=0 and dydx=4?x=0;y=0 and dydx=4?

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Expert Answer

Gennenzip
Answered 2021-06-26 Author has 96 answers

The given differential is d2ydx22dydx+10y=0 where x=0;y=0 and dydx=4.
The auxilary equation is given by m22m+10=0
m=2+(440)m=2±2(110)m=1±110m=1±9m=1±3i
Therefore the general olution is given by y(x)=ex(c1cos(3x)+c2sin(3x))
When x=0,y=0 therefore c1=0. Therefore y=c2ex(sin(3x))dy/x=c2ex(sin(3x)+3cos(3x))
Again when x=0,dydx=4, this implies that 4=3e2
Therefore the solution is y=(4ex3)sin(3x)

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