Question

Solve the following Cauchy-Euler equation \[ x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0, y(2)=-1, y^{\prime}(2)=4 \]

Answer 1

To solve the Cauchy-Euler equation \[ x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0 \] we can assume a solution of the form
`\displaystyle\sf y(x) = x^(r)`.

First, let's find the derivatives of
`\displaystyle\sf y(x)`:

`\displaystyle\sf y'(x) = r x^(r-1)`

`\displaystyle\sf y''(x) = r(r-1) x^(r-2)`

Now, we substitute these expressions back into the original equation:

`\displaystyle\sf x^(2) y'' + 2xy' - 6y = 0`

`\displaystyle\sf x^(2)(r(r-1) x^(r-2)) + 2x(r x^(r-1)) - 6(x^(r)) = 0`

Simplifying the equation:

`\displaystyle\sf r(r-1) x^(r) + 2rx^(r) - 6x^(r) = 0`

`\displaystyle\sf r(r-1) + 2r - 6 = 0`

`\displaystyle\sf r^(2) - r + 2r - 6 = 0`

`\displaystyle\sf r^(2) + r - 6 = 0`

Now we can solve this quadratic equation for
`\displaystyle\sf r`:

`\displaystyle\sf (r - 2)(r + 3) = 0`

This gives us two possible values for
`\displaystyle\sf r`:

`\displaystyle\sf r_(1) = 2`

`\displaystyle\sf r_(2) = -3`

Therefore, the general solution to the Cauchy-Euler equation is given by:

`\displaystyle\sf y(x) = C_(1) x^(2) + C_(2) x^(-3)`

To find the particular solution that satisfies the initial conditions
`\displaystyle\sf y(2) = -1` and
`\displaystyle\sf y'(2) = 4`, we substitute these values into the general solution and solve for the constants
`\displaystyle\sf C_(1)` and
`\displaystyle\sf C_(2)`.

Substituting
`\displaystyle\sf x = 2` and
`\displaystyle\sf y = -1`:

`\displaystyle\sf -1 = C_(1) (2^(2)) + C_(2) (2^(-3))`

`\displaystyle\sf -1 = 4C_(1) + (C_(2))/(8)`

Substituting
`\displaystyle\sf x = 2` and
`\displaystyle\sf y' = 4`:

`\displaystyle\sf 4 = 2C_(1) - 3C_(2)`

Now we have a system of linear equations that we can solve. Multiplying the first equation by 8 to eliminate fractions:

`\displaystyle\sf -8 = 32C_(1) + C_(2)`

Adding this

equation to the second equation:

`\displaystyle\sf -8 + 4 = 32C_(1) + C_(2) + 2C_(1) - 3C_(2)`

`\displaystyle\sf -4 = 34C_(1) - 2C_(2)`

Solving for
`\displaystyle\sf C_(1)`:

`\displaystyle\sf 34C_(1) - 2C_(2) = -4`

`\displaystyle\sf 34C_(1) = -4 + 2C_(2)`

`\displaystyle\sf C_(1) = (-4 + 2C_(2))/(34)`

Substituting this value of
`\displaystyle\sf C_(1)` back into the first equation:

`\displaystyle\sf -1 = 4C_(1) + (C_(2))/(8)`

`\displaystyle\sf -1 = 4\left((-4 + 2C_(2))/(34)\right) + (C_(2))/(8)`

`\displaystyle\sf -1 = (-16 + 8C_(2))/(34) + (C_(2))/(8)`

`\displaystyle\sf -1 = (-16 + 8C_(2))/(34) + (17C_(2))/(8)`

`\displaystyle\sf -1 = (-16(2) + 8C_(2)(2) + 17C_(2)(17))/(34(8))`

`\displaystyle\sf -1 = (-32 + 16C_(2) + 289C_(2))/(272)`

`\displaystyle\sf -1 = (305C_(2) - 32)/(272)`

`\displaystyle\sf -1(272) = 305C_(2) - 32`

`\displaystyle\sf -272 = 305C_(2) - 32`

`\displaystyle\sf -272 + 32 = 305C_(2)`

`\displaystyle\sf -240 = 305C_(2)`

`\displaystyle\sf C_(2) = (-240)/(305)`

`\displaystyle\sf C_(2) \approx -0.7877`

Substituting this value of
`\displaystyle\sf C_(2)` back into the equation for
`\displaystyle\sf C_(1)`:

`\displaystyle\sf C_(1) = (-4 + 2C_(2))/(34)`

`\displaystyle\sf C_(1) = (-4 + 2(-0.7877))/(34)`

`\displaystyle\sf C_(1) \approx -0.0646`

Therefore, the particular solution that satisfies the initial conditions is:

`\displaystyle\sf y(x) \approx -0.0646x^(2) - 0.7877x^(-3)`