Question

(d^2+5d-6)y=sin4x sinx​

Answer 1

Answer:

y = Ae^x + Be^-6x + 1/2[sin 3x - cos 3x/30 + 31cos5x-25sin5x/1586]

Step-by-step explanation:

Answer 2

$y = C_1 e^{-6x} + C_2 e^{x} + \frac{1}{2} \Big[ \frac{sin3x - cos3x}{30} + \frac{31cos5x - 25 sin5x}{1586} \Big]$

Step-by-step explanation:

Given: The differential equation $(D^{2} + 5D - 6) y = sin4x \ sinx$

To Find: Solution of the given differential equation

Solution:

• Finding Complementary Function (C.F.)

For the given differential equation, $(D^{2} + 5D - 6) y = sin4x \ sinx$, consider,

$m^{2} + 5m - 6 = 0$

$\Rightarrow (m+6) (m-1) = 0$

$\Rightarrow m = -6, 1$

Therefore, $\text{C.F.} = C_1 e^{-6x} + C_2 e^{x} \ \cdots \cdots (1)$

• Finding Particular Integral (P.I.)

To find the particular integral, consider the given differential equation such that we have,

$\text{P.I.} = \frac{1}{(D^{2} + 5D - 6) } sin4x \ sinx$

$\Rightarrow \text{P.I.} = \frac{1}{(D^{2} + 5D - 6) } \frac{(cos3x - cos5x)}{2} \because sinAsinB = \frac{cos(A-B) - cos(A+B)}{2}$

$\Rightarrow \text{P.I.} = \frac{1}{2} \frac{1}{(D^{2} + 5D - 6) } cos3x - \frac{1}{2} \frac{1}{(D^{2} + 5D - 6) } cos5x$

Now, consider $P.I. (1) = \frac{1}{2} \frac{1}{(D^{2} + 5D - 6) } cos3x$

$\Rightarrow P.I. (1) = \frac{1}{2} \frac{1}{(-(3)^{2} + 5D - 6) } cos3x \ \because \frac{1}{f(D^{2} )} cosax = \frac{1}{f(-a^{2} )} cosax$

$\Rightarrow P.I. (1) = \frac{1}{10} \frac{1}{(D -3)} cos3x$

To get $D^{2}$ at the denominator to apply the rule $\frac{1}{f(D^{2} )} cosax = \frac{1}{f(-a^{2} )} cosax$, multiply and divide $(D + 3)$ to get,

$\Rightarrow P.I. (1) = \frac{1}{10} \frac{1}{(D -3)}\times \frac{(D + 3)}{(D + 3)} cos3x$

$\Rightarrow P.I. (1) = \frac{1}{10} \frac{(D + 3)}{(D^2 - 3^2)} cos3x = \frac{1}{10} \frac{(D + 3)}{(-18)} cos3x$

$\Rightarrow P.I. (1) = \frac{1}{60} (sin3x - cos3x)$

Now, consider $P.I. (2) = \frac{1}{2} \frac{1}{(D^{2} + 5D - 6) } cos5x$

$\Rightarrow P.I. (2) = \frac{1}{2} \frac{1}{(-(5)^{2} + 5D - 6) } cos5x \ \because \frac{1}{f(D^{2} )} cosax = \frac{1}{f(-a^{2} )} cosax$

$\Rightarrow P.I. (2) = \frac{1}{2} \frac{1}{(5D -31)} cos5x$

To get $D^{2}$ at the denominator to apply the rule $\frac{1}{f(D^{2} )} cosax = \frac{1}{f(-a^{2} )} cosax$, multiply and divide $(5D + 31)$ to get,

$\Rightarrow P.I. (2) = \frac{1}{2} \frac{1}{(5D -31)}\times \frac{(5D + 31)}{(5D + 31)} cos5x$

$\Rightarrow P.I. (2) = \frac{1}{2} \frac{(5D + 31)}{(25D^2 - 31^2)} cos5x = \frac{1}{2} \frac{(5D + 31)}{(1586)} cos5x$

$\Rightarrow P.I. (2) = \frac{1}{2} \Big( \frac{25sin5x - 31cos5x}{1586} \Big)$

Therefore, $P.I. = P.I. (1) + P.I. (2) = \frac{1}{2} \Big[ \frac{sin3x - cos3x}{30} + \frac{31cos5x - 25 sin5x}{1586} \Big] \ \cdots \cdots (2)$

• Complete solution y = C.F. + P.I.

The solution is the sum of C.F. and P.I. Therefore, $y = C.F. + P.I. = C_1 e^{-6x} + C_2 e^{x} + \frac{1}{2} \Big[ \frac{sin3x - cos3x}{30} + \frac{31cos5x - 25 sin5x}{1586} \Big]$

Hence, the solution of the given differential equation is $y = C_1 e^{-6x} + C_2 e^{x} + \frac{1}{2} \Big[ \frac{sin3x - cos3x}{30} + \frac{31cos5x - 25 sin5x}{1586} \Big]$