Question

Find the general solution of $$ (x+1)^{2} \frac{d^{2} y}{d x^{2}}+3(x+1) \frac{d y}{d x}+y=x^{2}. $$

Short answer

The general solution is:\[ y = \frac{C_1 + C_2 \text{ln}(x + 1)}{x + 1} + \frac{x^2}{3} - x - \frac {5}{3}.\]

Step-by-step solution

- Identify the type of differential equation

Rewrite the given equation: \[(x+1)^{2} \frac{d^{2} y}{d x^{2}} + 3(x+1) \frac{d y}{d x} + y = x^{2}.\] This is a second-order linear non-homogeneous differential equation with variable coefficients of the form: \[a(x) \frac{d^{2} y}{d x^{2}} + b(x) \frac{d y}{d x} + c(x)y = f(x).\] Here, \[a(x) = (x+1)^2, b(x) = 3(x+1), c(x) = 1, \text{and} \, f(x) = x^2.\]

- Find the complementary solution

Solve the homogeneous part of the equation:\[(x+1)^{2} \frac{d^{2} y}{d x^{2}} + 3(x+1) \frac{d y}{d x} + y = 0.\] Let \[t = x + 1\], then the equation becomes:\[t^2 \frac{d^{2} y}{d t^{2}} + 3t \frac{d y}{d t} + y = 0.\] Assume a solution of the form \[y = t^m\]. Substituting \[y = t^m\] into the homogeneous equation gives the characteristic equation:\[t^2 m(m-1)t^{m-2} + 3 t m t^{m-1} + t^m = 0,\] simplify to get the characteristic equation:\[m(m - 1) + 3m + 1 = 0\], this is:\[m^2 + 2m + 1 = 0\], solving the quadratic equation, we get:\[(m + 1)^2 = 0\], thus \[m = -1 \text{(double root)}\]. So the complementary solution is:\[y_c = (C_1 + C_2 \text{ln}(t))/t,\] reverting back to variable x:\[y_c = \frac{C_1 + C_2 \text{ln}(x + 1)}{x + 1}.\]

- Find the particular solution

Assume a form for the particular solution given \[f(x) = x^2\]. Since \[f(x)\] is a polynomial of order 2, assume the particular solution to be :\[y_p = Ax^2 + Bx + C\]. Find the first and second derivatives:\[\frac{d y_p}{d x} = 2Ax + B\] \[\frac{d^{2} y_p}{d x^{2}} = 2A\]. Substitute these into the original differential equation:\[(x+1)^{2}(2A) + 3(x+1)(2Ax + B) + (Ax^2 + Bx + C) = x^2\]. Simplifying this gives:\[2A(x+1)^2 + 3(2Ax + B)(x+1) + Ax^2 + Bx + C = x^2\]. Expand and collect like terms of power x, equate the coefficients to find A, B, C:\[2 Ax^2+4Ax+2A + 6Ax+3Bx + 3B + Ax^2 + Bx + C = x^2\], resulting in:\[ (2A + A)x^2 + (4A + 6A + B)x + so we get\ C +(2A + 3B) = 0,\ A = 1/3 , B = -1 , and\ -2A + C + 3B -1)\] Thus the particular solution is: \[ y_p = \frac{x^2}{3} - x - \frac {5}{3}\]

- Write the general solution

Combine the complementary solution \[y_c\] and particular solution \[y_p\] to find the general solution:\[y = y_c + y_p \]\[ y = \frac{C_1 + C_2 \text{ln}(x + 1)}{x + 1} + \frac{x^2}{3} - x - \frac {5}{3}\].

Key concepts

Complementary Solution

When solving second-order linear differential equations, we start by finding the complementary solution. This involves solving the homogeneous version of the equation, where the right-hand side is zero. For instance, with our equation \[(x+1)^{2} \frac{d^{2} y}{d x^{2}} + 3(x+1) \frac{d y}{d x} + y = 0,\]we assume a solution form such as \(y = t^m\). By substituting and simplifying, we derive the characteristic equation, which in this case is \(m^2 + 2m + 1 = 0\). Solving this quadratic gives double roots of -1, leading to a complementary solution of the form \[ y_c = \frac{C_1 + C_2 \text{ln}(x + 1)}{x + 1}. \]This method helps identify the structure of solutions without considering the non-homogeneous part.

Particular Solution

Next, we find the particular solution that accounts for the non-homogeneous part of the differential equation. Given that our original equation includes \(f(x) = x^2\), a polynomial of degree 2, we assume a particular solution of the form \[ y_p = Ax^2 + Bx + C. \]By deriving the first and second derivatives and substituting them back into the original equation, we simplify and equate the coefficients. Through this process, we determine the constants \(A\), \(B\), and \(C\). Our particular solution then becomes \[ y_p = \frac{x^2}{3} - x - \frac{5}{3}. \]This step ensures that we have a solution that fits the specific non-homogeneous equation.

Characteristic Equation

The characteristic equation is fundamental when solving the homogeneous part of a second-order linear differential equation. This equation is derived from assuming a solution form like \(y = t^m\), which transforms the differential equation into one that depends only on \(m\). In our example, replacing \(y = t^m\) into \[ t^2 (m(m-1)t^{m-2}) + 3t (m t^{m-1}) + t^m = 0, \]and simplifying, we obtain \[ m(m - 1) + 3m + 1 = 0. \]This simplifies further to \[ m^2 + 2m + 1 = 0, \]solving which provides the roots \(m = -1\) (a double root). The characteristic equation thus informs the nature of the solutions, whether they be exponentially decaying, oscillatory, or otherwise.

Non-Homogeneous Differential Equations

Non-homogeneous differential equations include a term on the right-hand side that isn't zero, often representing an external force or input. For our exercise, the equation \[ (x+1)^{2} \frac{d^{2} y}{d x^{2}} + 3(x+1) \frac{d y}{d x} + y = x^2, \]is non-homogeneous due to the \(x^2\) term. Solving such equations typically involves two main steps: finding the complementary solution for the corresponding homogeneous equation and then finding the particular solution for the entire equation. Finally, combining these solutions gives the general solution, which effectively captures the behavior dictated by both the inherent properties of the system and the external input or forces.