Questions: Find a general solution to the given equation for t<0. y''(t) - (3/t) y'(t) + (29/t^2) y(t) = 0

Find a general solution to the given equation for t<0.

y''(t) - (3/t) y'(t) + (29/t^2) y(t) = 0
Transcript text: Find a general solution to the given equation for $\mathrm{t}<0$. \[ y^{\prime \prime}(t)-\frac{3}{t} y^{\prime}(t)+\frac{29}{t^{2}} y(t)=0 \]
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Solution

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Solution Steps

To solve the given differential equation, we recognize it as a Cauchy-Euler equation. The general approach involves assuming a solution of the form y(t)=tm y(t) = t^m and substituting it into the differential equation to find the characteristic equation. Solving the characteristic equation will give us the values of m m , which we use to construct the general solution.

Step 1: Characteristic Equation

We start with the characteristic equation derived from the assumption y(t)=tm y(t) = t^m : m(m1)3m+29=0 m(m - 1) - 3m + 29 = 0 This simplifies to: m24m+29=0 m^2 - 4m + 29 = 0

Step 2: Solve for m m

Using the quadratic formula m=b±b24ac2a m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , we find the roots: m=4±(4)2412921=4±161162=4±1002 m = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 29}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 116}}{2} = \frac{4 \pm \sqrt{-100}}{2} This gives us: m=2±5i m = 2 \pm 5i

Step 3: General Solution

Since the roots are complex, the general solution for the differential equation is given by: y(t)=t2(C1cos(5lnt)+C2sin(5lnt)) y(t) = t^2 \left( C_1 \cos(5 \ln |t|) + C_2 \sin(5 \ln |t|) \right) where C1 C_1 and C2 C_2 are arbitrary constants.

Final Answer

The general solution is: y(t)=t2(C1cos(5lnt)+C2sin(5lnt)) \boxed{y(t) = t^2 \left( C_1 \cos(5 \ln |t|) + C_2 \sin(5 \ln |t|) \right)}

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