Math Problem Statement

x^2y''+2xy'-42y=0, y(1)=-8,y'(1)=0

Solution

We are given the second-order linear differential equation:

x2y+2xy42y=0x^2 y'' + 2x y' - 42 y = 0

with initial conditions: y(1)=8,y(1)=0.y(1) = -8, \quad y'(1) = 0.

Step 1: Solve the homogeneous differential equation

This is a Cauchy-Euler (or Euler-Cauchy) equation of the form: x2y+axy+by=0.x^2 y'' + a x y' + b y = 0. In our case, a=2a = 2 and b=42b = -42.

The general solution to a Cauchy-Euler equation can be found by assuming a solution of the form: y(x)=xr,y(x) = x^r, where rr is a constant. We differentiate y(x)y(x) with respect to xx: y(x)=rxr1,y(x)=r(r1)xr2.y'(x) = r x^{r-1}, \quad y''(x) = r(r-1) x^{r-2}.

Step 2: Substitute into the differential equation

Substitute y(x)=xry(x) = x^r, y(x)=rxr1y'(x) = r x^{r-1}, and y(x)=r(r1)xr2y''(x) = r(r-1) x^{r-2} into the original equation: [ x^2 [r(r-1) x^{r-2}] + 2x [r x^{r-1}] - 42 x^r = 0. ] Simplify each term: r(r1)xr+2rxr42xr=0.r(r-1) x^r + 2r x^r - 42 x^r = 0. Factor out xrx^r: xr[r(r1)+2r42]=0.x^r [r(r-1) + 2r - 42] = 0. We are left with the characteristic equation: r(r1)+2r42=0.r(r-1) + 2r - 42 = 0. Simplify: r2+r42=0.r^2 + r - 42 = 0.

Step 3: Solve the characteristic equation

Solve the quadratic equation r2+r42=0r^2 + r - 42 = 0 using the quadratic formula: r=1±124(1)(42)2(1)=1±1+1682=1±1692=1±132.r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-42)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 168}}{2} = \frac{-1 \pm \sqrt{169}}{2} = \frac{-1 \pm 13}{2}. The two roots are: r1=1+132=6,r2=1132=7.r_1 = \frac{-1 + 13}{2} = 6, \quad r_2 = \frac{-1 - 13}{2} = -7.

Step 4: General solution

The general solution to the differential equation is: y(x)=C1x6+C2x7.y(x) = C_1 x^6 + C_2 x^{-7}.

Step 5: Apply initial conditions

We now use the initial conditions to determine C1C_1 and C2C_2.

  • Initial condition 1: y(1)=8y(1) = -8

Substitute x=1x = 1 into the general solution: y(1)=C1(1)6+C2(1)7=C1+C2=8.y(1) = C_1 (1)^6 + C_2 (1)^{-7} = C_1 + C_2 = -8. Thus, we have: C1+C2=8.(1)C_1 + C_2 = -8. \quad (1)

  • Initial condition 2: y(1)=0y'(1) = 0

First, differentiate y(x)y(x): y(x)=6C1x57C2x8.y'(x) = 6C_1 x^5 - 7C_2 x^{-8}. Substitute x=1x = 1 into the derivative: y(1)=6C1(1)57C2(1)8=6C17C2=0.y'(1) = 6C_1 (1)^5 - 7C_2 (1)^{-8} = 6C_1 - 7C_2 = 0. Thus, we have: 6C17C2=0.(2)6C_1 - 7C_2 = 0. \quad (2)

Step 6: Solve the system of equations

We now solve the system of equations:

  1. C1+C2=8C_1 + C_2 = -8
  2. 6C17C2=06C_1 - 7C_2 = 0

From equation (2), solve for C1C_1: C1=76C2.C_1 = \frac{7}{6} C_2. Substitute this into equation (1): 76C2+C2=8,\frac{7}{6} C_2 + C_2 = -8, 136C2=8,\frac{13}{6} C_2 = -8, C2=8×613=4813.C_2 = -8 \times \frac{6}{13} = -\frac{48}{13}.

Now, substitute C2=4813C_2 = -\frac{48}{13} into C1=76C2C_1 = \frac{7}{6} C_2: C1=76×(4813)=7×486×13=33678=16839.C_1 = \frac{7}{6} \times \left(-\frac{48}{13}\right) = -\frac{7 \times 48}{6 \times 13} = -\frac{336}{78} = -\frac{168}{39}.

Final solution:

Thus, the solution to the differential equation is: y(x)=16839x64813x7.y(x) = -\frac{168}{39} x^6 - \frac{48}{13} x^{-7}.

Would you like further details or explanations?

Related Questions:

  1. What is a Cauchy-Euler equation, and when is it used?
  2. How is the general solution for a Cauchy-Euler equation derived?
  3. Can the method used here apply to non-homogeneous equations?
  4. What would change if different initial conditions were given?
  5. How does the quadratic formula help in solving second-order differential equations?

Tip: To solve higher-order differential equations, always try to reduce them to familiar forms like Cauchy-Euler, which simplifies the solution process.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Cauchy-Euler Equations
Quadratic Equations

Formulas

Cauchy-Euler form: x^2y'' + axy' + by = 0
Characteristic equation: r^2 + (a-1)r + b = 0
Quadratic formula: r = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula
Cauchy-Euler Equation Solution Method

Suitable Grade Level

Undergraduate