Question

solve (2x+3)^2d^2y/dx^2-(2x+3)dy/dx-12y=6x

Answer 1

Answer on Question #81140 – Math – Differential Equations

Question

Solve

$$(2x + 3)^2 \frac{d^2 y}{d x^2} - 2(2x + 3) \frac{d y}{d x} - 12y = 6x$$

Solution

Given differential equation is a Legendre’s differential equation.

Let $2x + 3 = e^z \Rightarrow z = \ln(2x + 3)$

$$x = \frac{e^z - 3}{2}$$ $$\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = \frac{2}{2x + 3} \cdot \frac{dy}{dz}$$ $$(2x + 3) \frac{dy}{dx} = 2 \frac{dy}{dz} = 2Dy, \quad D = \frac{d}{dz}$$ $$(2x + 3)^2 \frac{d^2 y}{d x^2} = 2^2 D(D - 1)y$$

Substitute

$$(2^2 D(D - 1) - 2 \cdot 2D - 12)y = 6\left(\frac{e^z - 3}{2}\right)$$ $$(4D^2 - 4D - 4D - 12)y = 3(e^z - 3)$$ $$(4D^2 - 8D - 12)y = 3(e^z - 3)$$

We get a linear differential equation with constant coefficients.

$$4D^2 - 8D - 12 = 0$$ $$D^2 - 2D - 3 = 0$$ $$D = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = 1 \pm 2$$ $$D = 3, -1$$ $$y_C = c_1 e^{3z} + c_2 e^{-z} = c_1 (2x + 3)^3 + c_2 (2x + 3)^{-1}$$

General solution

$$y = y_C + Y_P$$ $$\begin{aligned} Y_P &= \frac{1}{4D^2 - 8D - 12} \cdot 3(e^z - 3) \\ &= \frac{3}{4} \cdot \frac{1}{D^2 - 2D - 3} e^z - \frac{9}{4} \cdot \frac{1}{D^2 - 2D - 3} e^{0 \cdot z} \\ &= -\frac{3}{16} e^z + \frac{3}{4} = -\frac{3}{16} (2x + 3) + \frac{3}{4} \end{aligned}$$

Therefore, general solution

$$y = c_1 (2x + 3)^3 + c_2 (2x + 3)^{-1} - \frac{3}{16} (2x + 3) + \frac{3}{4}.$$

Answer: $y = c_{1}(2x + 3)^{3} + c_{2}(2x + 3)^{-1} - \frac{3}{16}(2x + 3) + \frac{3}{4}$.

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