Use the method of reduction of order to find A second solution of the given DE y'' + p(t)y' + q(t)y = 0 t^2y'' + 2ty' - 2y = 0 t>0 y(1) = t let y = v(t)y(t) => y = vt y' = v't + tv y'' = v''t + tv' + tv' + v't = v''t + 2v't + tv' t^2(v''t + 2v't + tv') + 2t(v't + v) - 2(vt) = 0 t^3v'' + 2t^2v' + t^2v + 2tv't + 2tv - 2vt = 0 t^3v'' + 2t^2v' + 2v't - 2v't + 3v = 0 t^2(tv'' + 2v') + 2t(v' - v) + 3v = 0
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The method of reduction of order (Section 3.4) can also be used for the nonhomogeneous equation y'' + p(t)y' + q(t)y = g(t), (38) provided one solution y1 of the corresponding homogeneous equation is known. Let y = v(t)y1(t) and show that y satisfies equation 38 if v is a solution of y1(t)v'' + (2y1'(t) + p(t)y1(t))v' = g(t). (39) Equation 39 is a first order linear equation for v'. Solving this equation, integrating the result, and then multiplying by y1(t) leads to the general solution of the first equation. Use the method above to solve the differential equation t^2y'' - 2ty' + 2y = 4t^2, t > 0, y1(t) = t. NOTE: Use C1, C2, ... for the constants of integration. Y(t) =
Sri K.
The differential equation y'' - t(t+2)y + (t+2)y = 0 has Y1 as a solution. Applying reduction of order, we set Y2 = v.Y1 = V. Then (using the prime notation for the derivatives) y2 = Y1'. So, substituting Y2 and its derivatives into the left side of the differential equation and reducing, we get 2Y1(t + 2)y2 + (t + 2)y2. The reduced form has a common factor of t^3 which we can divide out of the equation. Since this equation does not have any V terms in it, we can make the substitution u, giving us the first-order linear equation in u. If we use C as the constant of integration, the solution to this equation is u = ∫(t+2)dt. Integrating to get V, and then finding Y2 gives the general solution: CY1 + CY2.
Adi S.
The method of reduction of order (Section 3.4) can also be used for the nonhomogeneous equation y'' + p(t)y' + q(t)y = g(t), (38) provided one solution y1 of the corresponding homogeneous equation is known. Let y = v(t)y1(t) and show that y satisfies equation 38 if v is a solution of y1(t)v'' + (2y1'(t) + p(t)y1(t))v' = g(t). (39) Equation 39 is a first order linear equation for v'. Solving this equation, integrating the result, and then multiplying by y1(t) leads to the general solution of the first equation. Use the method above to solve the differential equation t^2y'' - 2ty' + 2y = 11t^2, t > 0, y1(t) = t. NOTE: Use c1, c2, ... for the constants of integration. Y(t) =
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