Question

Q. solve (D3−7D−6)y=1+x

Answer 1

Answer on Question #64772 – Math – Differential Equations Question

Solve

$$y^{\prime \prime \prime} - 7 y^{\prime} - 6 y = 1 + x.$$

Solution

Equation

$$(D^3 - 7D - 6)y = 1 + x,$$

that is,

$$y^{\prime \prime \prime} - 7 y^{\prime} - 6 y = 1 + x$$

is a linear non-homogeneous ordinary differential equation of the third order.

First, we solve the corresponding characteristic equation:

$$\begin{array}{l} k^3 - 7k - 6 = 0, \\ (k + 1)(k^2 - k - 6) = 0, \\ (k + 1)(k + 2)(k - 3) = 0, \\ k = -1, k = -2, k = 3. \end{array}$$

Then the solution of the homogeneous differential equation is the complementary function

$$y_c(x) = c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{3x},$$

where $c_1, c_2$, and $c_3$ are any real constants.

The additional solution to the complementary function is a particular integral, which we looking for in the form of

$$y_p(x) = a x + b.$$

To find the unknown constants $a$ and $b$ we substitute $y_p$ into the original equation:

$$\begin{array}{l} (a x + b)^{\prime \prime \prime} - 7(a x + b)^{\prime} - 6(a x + b) = 1 + x, \\ -7a - 6b - 6a x = 1 + x. \end{array}$$

Now we equate the coefficients of equal powers of $x$:

$$-7a - 6b = 1$$

and

$$-6a = 1,$$

whence

$$a = -\frac{1}{6}, \text{ and } b = \frac{1}{36}.$$

So we have

$$y_p(x) = -\frac{1}{6} x + \frac{1}{36}.$$

Finally, the general solution to the given differential equation can be written as

$$y(x) = y_c(x) + y_p(x) = c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{3x} - \frac{1}{6} x + \frac{1}{36}.$$

Answer: $y(x) = c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{3x} - \frac{1}{6} x + \frac{1}{36}$.

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