# Solve the differential equation 2y"+20y'+51y=0, y(0)=2 y'(0)=0

Question
Differential equations
Solve the differential equation $$2y"+20y'+51y=0, y(0)=2$$
$$y'(0)=0$$

2021-02-09
Step 1
Let $$y=e^{mt}$$ be a trial solution. Then the auxiliary equation is given by $$2m^{2}+20m+51=0 as e^{mt}\neq0$$, for all t
Thus the solutions of auxiliary equations are
$$m= -5\pm i\sqrt{\frac{2}{2}}$$
Therefore the solution of the differential equation is given by
$$y= e^{-5}(c_{1}\cos(\frac{t}{\sqrt 2})+c_{2}\\sin(\frac{t}{\sqrt 2}))$$
Now apply initial condition
$$y(0)= e^{-5\times0}(c_{1}\cos(\frac{0}{\sqrt 2})+c_{2}\\sin(\frac{0}{\sqrt 2}))= 1\times(\cos(0)c_{1}+\\sin(0)c_{2})=c_{1}$$
Thus y(0)=2 gives us $$c_{1}=2$$
Step 2
Further by allpying the product rule of differention we get
$$y'(x)=(e^{-5t})(c_{1}\cos(\frac{t}{\sqrt 2})+c_{2}\\sin(\frac{t}{\sqrt 2}))+(c_{1}(\frac{t}{\sqrt 2})+c_{2}\sin(\frac{t}{\sqrt 2}))' e^{-5t}$$
$$=(-5e^{-5t})(c_{1}\cos(\frac{t}{\sqrt 2})+c_{2}\sin(\frac{t}{\sqrt 2}))+(-c_{1}\sin(\frac{t}{\sqrt 2})+c_{2}cos(\frac{t}{\sqrt 2}))\sqrt 2 e^{-5t}$$
$$= -5e^{-5t}(c_{1}\cos(\frac{t}{\sqrt 2})+c_{2}\sin(\frac{t}{\sqrt 2}))+(e^{-5t}(-c_{1}\sin(\frac{t}{\sqrt 2})+c_{2}\cos(\frac{t}{\sqrt 2}))\sqrt 2$$
Notice that
$$y'(0)= -5e^{-5*0}(c_{1}\cos(\frac{0}{\sqrt 2})+c_{2}\sin(\frac{0}{\sqrt 2}))+(e^{-5*0}(-c_{1}\sin(\frac{0}{\sqrt 2})+c_{2}\cos(\frac{0}{\sqrt 2}))\sqrt 2$$
$$= -5e^{5*0}(\cos(0)c_{1}+\sin(0)c_{2})+(e^{-5*0}(\cos(0)c_{2}-\sin(0)c_{1}))\sqrt 2= -5c_{1}+c_{2}\sqrt 2$$
Now y'(0)=0 gives us
$$-5c_{1}+c_{2}\sqrt 2=0 \implies -10+c_{2}\sqrt 2=0 \ [\because c_{1}=2]$$
$$\implies c_{2}=10\sqrt 2$$
Therefore the required solution is given by
$$y= e^{-5t}(2\cos(\frac{t}{\sqrt 2})+10\sqrt 2\sin(\frac{t}{\sqrt 2}))$$

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