Question

(2x+1)^2 d^2y/dx^2 - 2(2x+1) dy/dx- 12y= 6x
solve by taking substitution (2x+1)= e^t

Answer 1

Answer on Question #80967 – Math – Differential Equations

Question

$$2x + 1)^2 \frac{d^2 y}{dx^2} - 2(2x + 1) \frac{dy}{dx} - 12y = 6x$$

solve by taking substitution $(2x + 1) = e^t$

Solution

Consider

$$2x + 1 = e^t$$

We can calculate $x, t$ and the corresponding derivatives as

$$\begin{array}{l} x = \frac{e^t - 1}{2} \\ t = \log(2x + 1) \\ \frac{dt}{dx} = \log(2x + 1)' = \frac{2}{2x + 1} \\ \frac{d^2 t}{dx^2} = \log(2x + 1)' = \frac{-4}{(2x + 1)^2} \\ \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{2}{2x + 1} \frac{dy}{dt} \\ \frac{d^2 y}{dx^2} = \frac{d^2 t}{dx^2} \frac{dy}{dt} + \left(\frac{dt}{dx}\right)^2 \frac{d^2 y}{dt^2} = \frac{-4}{(2x + 1)^2} \frac{dy}{dt} + \frac{4}{(2x + 1)^2} \frac{d^2 y}{dt^2} \\ \end{array}$$

Then the initial equation can be written as:

$$\begin{array}{l} e^{2t} \left(\frac{-4}{e^{2t}} \frac{dy}{dt} + \frac{4}{e^{2t}} \frac{d^2 y}{dt^2}\right) - 2e^t \frac{2}{e^t} \frac{dy}{dt} - 12y = 3e^t - 3 \\ 4 \frac{d^2 y}{dt^2} - 8 \frac{dy}{dt} - 12y = 3e^t - 3 \\ \end{array}$$

The general solution for this equation will be the sum of the complementary solution and a particular solution.

We find the complementary solution by solving

$$4 \frac{d^2 y}{dt^2} - 8 \frac{dy}{dt} - 12y = 0$$

Let's assume a solution will be proportional to $e^{at}$ for some constant $a$

Substitute $y = e^{at}$ into the differential equation:

$$\begin{array}{l} 4 \frac{d^2 (e^{at})}{dt^2} - 8 \frac{d (e^{at})}{dt} - 12 (e^{at}) = 0 \\ 4a^2 e^{at} - 8a e^{at} - 12 e^{at} = 0 \\ \end{array}$$

Since $e^{at} \neq 0$ for any finite $a$, the zeros must come from the polynomial:

$$4a^2 - 8a - 12 = 0$$

Factor

$$4(a - 3)(a + 1) = 0$$

Thus, the roots are

$$a_1 = -1$$ $$a_2 = 3$$

The root $a_1 = -1$ gives $y_1 = C_1 e^{-t}$ as a solution, where $C_1$ is an arbitrary constant.

The root $a_2 = 3$ gives $y_2 = C_2 e^{3t}$ as a solution, where $C_2$ is an arbitrary constant.

The general solution is the sum of the above solutions:

$$y = C_1 e^{-t} + C_2 e^{3t}$$

We determine a particular solution to $4\frac{d^2y}{dt^2} - 8\frac{dy}{dt} - 12y = 3e^t - 3$ by the method of undetermined coefficients:

A particular solution will be the sum of particular solutions to $4\frac{d^2y}{dt^2} - 8\frac{dy}{dt} - 12y = -3$ and $4\frac{d^2y}{dt^2} - 8\frac{dy}{dt} - 12y = 3e^t$

A particular solution to $4\frac{d^2y}{dt^2} - 8\frac{dy}{dt} - 12y = -3$ is of the form $y_3 = b_1$

A particular solution to $4\frac{d^2y}{dt^2} - 8\frac{dy}{dt} - 12y = 3e^t$ is of the form $y_4 = b_2 e^t$

Thus, the total particular solution can be found as

$$y_p = b_1 + b_2 e^t$$

Substituting

$$4 \frac{d^2 y_p}{dt^2} - 8 \frac{dy_p}{dt} - 12 y_p = 3 e^t - 3$$

We get

$$4 b_2 e^t - 8 b_2 e^t - 12 (b_1 + b_2 e^t) = 3 e^t - 3$$

Simplifying

$$-12 b_1 - 16 b_2 e^t = -3 + 3 e^t$$

Equating the coefficients before the corresponding powers of $e^t$ we get

$$-12 b_1 = -3$$ $$-16 b_2 = 3$$

So

$$b_1 = 1/4$$ $$b_2 = \frac{-3}{16}$$

Now substituting to $y_p$ we get

$$y_p = \frac{1}{4} - \frac{3}{16} e^t$$

Now the general solution can be found as

$$y = C_1 e^{-t} + C_2 e^{3t} + \frac{1}{4} - \frac{3}{16} e^t$$

Now substituting $e^t = 2x + 1$ we get

$$y = \frac{C_1}{2x + 1} + C_2 (2x + 1)^3 + \frac{1}{4} - \frac{3}{16} (2x + 1)$$

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