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Y''+y'+y=0, y(0)=1, y'(0)=0

User Samir Adel
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3 votes

Answer:


y=e^{(-t)/(2)}\left ( \cos\left ( (√(3)t)/(2) \right )+(1)/(√(3))\sin \left ( (√(3)t)/(2) \right ) \right )

Explanation:

A second order linear , homogeneous ordinary differential equation has form
ay''+by'+cy=0.

Given:
y''+y'+y=0

Let
y=e^(rt) be it's solution.

We get,


\left ( r^2+r+1 \right )e^(rt)=0

Since
e^(rt)\\eq 0,
r^2+r+1=0

{ we know that for equation
ax^2+bx+c=0, roots are of form
x=(-b\pm √(b^2-4ac))/(2a) }

We get,


y=(-1\pm √(1^2-4))/(2)=(-1\pm √(3)i)/(2)

For two complex roots
r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form
y=e^(\alpha t)\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e
y=e^{(-t)/(2)}\left ( c_1\cos\left ( (√(3)t)/(2) \right )+c_2\sin \left ( (√(3)t)/(2) \right ) \right )

Applying conditions y(0)=1 on
e^{(-t)/(2)}\left ( c_1\cos\left ( (√(3)t)/(2) \right )+c_2\sin \left ( (√(3)t)/(2) \right ) \right ),
c_1=1

So, equation becomes
y=e^{(-t)/(2)}\left ( \cos\left ( (√(3)t)/(2) \right )+c_2\sin \left ( (√(3)t)/(2) \right ) \right )

On differentiating with respect to t, we get


y'=(-1)/(2)e^{(-t)/(2)}\left ( \cos\left ( (√(3)t)/(2) \right )+c_2\sin \left ( (√(3)t)/(2) \right ) \right )+e^{(-t)/(2)}\left ( (-√(3))/(2) \sin \left ( (√(3)t)/(2) \right )+c_2(√(3))/(2)\cos\left ( (√(3)t)/(2) \right )\right )

Applying condition: y'(0)=0, we get
0=(-1)/(2)+(√(3))/(2)c_2\Rightarrow c_2=(1)/(√(3))

Therefore,


y=e^{(-t)/(2)}\left ( \cos\left ( (√(3)t)/(2) \right )+(1)/(√(3))\sin \left ( (√(3)t)/(2) \right ) \right )

User Doug Smith
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