Question

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

Answer 1

`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 )`