Question

Solve differential equation :-
1- Y^4 +9 y″ =(x^2 +1 )sin3x
2- y″ +4 y = sin^2 x

Answer 1

Solve $9 \frac{d^2 y(x)}{dx^2} + y(x) = (x^2 + 1) \sin(3x)$:

The general solution will be the sum of

the complementary solution and particular solution.

Find the complementary solution by solving $9 \frac{d^2 y(x)}{dx^2} + y(x) = 0$:

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$.

Substitute $y(x) = e^{\lambda x}$ into the differential equation:

$$9 \frac{d^2}{dx^2} (e^{\lambda x}) + e^{\lambda x} = 0$$

Substitute $\frac{d^2}{dx^2} (e^{\lambda x}) = \lambda^2 e^{\lambda x}$:

$$9 \lambda^2 e^{\lambda x} + e^{\lambda x} = 0$$

Factor out $e^{\lambda x}$:

$$(9 \lambda^2 + 1) e^{\lambda x} = 0$$

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

$$9 \lambda^2 + 1 = 0$$

Solve for $\lambda$:

$$\lambda = \frac{i}{3} \quad \text{or} \quad \lambda = -\frac{i}{3}$$

The roots $\lambda = \pm \frac{i}{3}$ give $y_1(x) = c_1 e^{\frac{i x}{3}}$, $y_2(x) = c_2 e^{-\frac{i x}{3}}$ as solutions, where $c_1$ and $c_2$ are arbitrary constants.

The general solution is the sum of the above solutions:

$$y(x) = y_1(x) + y_2(x) = c_1 e^{\frac{i x}{3}} + c_2 e^{-\frac{i x}{3}}$$

Apply Euler's identity $e^{\alpha + i\beta} = e^{\alpha}\cos(\beta) + i e^{\alpha}\sin(\beta)$:

$$y(x) = c_1 \left(\cos\left(\frac{x}{3}\right) + i \sin\left(\frac{x}{3}\right)\right) + c_2 \left(\cos\left(\frac{x}{3}\right) - i \sin\left(\frac{x}{3}\right)\right)$$

Regroup terms:

$$y(x) = (c_1 + c_2) \cos\left(\frac{x}{3}\right) + i (c_1 - c_2) \sin\left(\frac{x}{3}\right)$$

Redefine $c_1 + c_2$ as $c_1$ and $i(c_1 - c_2)$

as $c_2$, since these are arbitrary constants:

$$y(x) = c_1 \cos\left(\frac{x}{3}\right) + c_2 \sin\left(\frac{x}{3}\right)$$

Determine the particular solution to $9 \frac{d^2 y(x)}{dx^2} + y(x) = (x^2 + 1) \sin(3x)$ by the method of undetermined coefficients:

The particular solution to $9 \frac{d^2 y(x)}{dx^2} + y(x) = (x^2 + 1) \sin(3x)$ is of the form:

$$\begin{array}{l} y_p(x) = a_1 \cos(3x) + a_2 x \cos(3x) + \\ a_3 x^2 \cos(3x) + a_4 \sin(3x) + a_5 x \sin(3x) + a_6 x^2 \sin(3x) \end{array}$$

Solve for the unknown constants $a_1, a_2, a_3, a_4, a_5$, and $a_6$:

Compute $\frac{d^2 y_p(x)}{dx^2}$:

$$\begin{array}{l} \frac{d^2 y_p(x)}{dx^2} = \frac{d^2}{dx^2} \left(a_1 \cos(3x) + a_2 x \cos(3x) + \right. \\ \left. a_3 x^2 \cos(3x) + a_4 \sin(3x) + a_5 x \sin(3x) + a_6 x^2 \sin(3x) \right) \\ = -9 a_1 \cos(3x) - 9 a_2 x \cos(3x) - 6 a_2 \sin(3x) + 2 a_3 \cos(3x) - \\ \left. 9 a_3 x^2 \cos(3x) - 12 a_3 x \sin(3x) - 9 a_4 \sin(3x) + 6 a_5 \cos(3x) - \right. \\ \left. 9 a_5 x \sin(3x) + 12 a_6 x \cos(3x) + 2 a_6 \sin(3x) - 9 a_6 x^2 \sin(3x) \right) \end{array}$$

Substitute the particular solution $y_{p}(x)$ into the differential equation:

$$9 \frac {d ^ {2} y _ {p} (x)}{d x ^ {2}} + y _ {p} (x) = \sin (3 x) + x ^ {2} \sin (3 x)$$ $$\begin{array}{l} 9 \left(- 9 a _ {1} \cos (3 x) - 9 a _ {2} x \cos (3 x) - 6 a _ {2} \sin (3 x) + 2 a _ {3} \cos (3 x) - \right. \\ \left. 9 a _ {3} x ^ {2} \cos (3 x) - 12 a _ {3} x \sin (3 x) - 9 a _ {4} \sin (3 x) + 6 a _ {5} \cos (3 x) - \right. \\ \left. 9 a _ {5} x \sin (3 x) + 12 a _ {6} x \cos (3 x) + 2 a _ {6} \sin (3 x) - 9 a _ {6} x ^ {2} \sin (3 x)\right) + \\ \left(a _ {1} \cos (3 x) + a _ {2} x \cos (3 x) + a _ {3} x ^ {2} \cos (3 x) + a _ {4} \sin (3 x) + \right. \\ \left. a _ {5} x \sin (3 x) + a _ {6} x ^ {2} \sin (3 x)\right) = \sin (3 x) + x ^ {2} \sin (3 x) \end{array}$$

Simplify:

$$\begin{array}{l} \left(- 80 a _ {1} + 18 a _ {3} + 54 a _ {5}\right) \cos (3 x) + \left(- 80 a _ {2} + 108 a _ {6}\right) x \cos (3 x) - \\ \left. 80 a _ {3} x ^ {2} \cos (3 x) + \left(- 54 a _ {2} - 80 a _ {4} + 18 a _ {6}\right) \sin (3 x) + \right. \\ \left(- 108 a _ {3} - 80 a _ {5}\right) x \sin (3 x) - 80 a _ {6} x ^ {2} \sin (3 x) = \sin (3 x) + x ^ {2} \sin (3 x) \end{array}$$

Equate the coefficients of $\cos (3x)$ on both sides of the equation:

$$- 80 a _ {1} + 18 a _ {3} + 54 a _ {5} = 0$$

Equate the coefficients of $x \cos (3x)$ on both sides of the equation:

$$- 80 a _ {2} + 108 a _ {6} = 0$$

Equate the coefficients of $x^{2} \cos (3x)$ on both sides of the equation:

$$- 80 a _ {3} = 0$$

Equate the coefficients of $\sin (3x)$ on both sides of the equation:

$$- 54 a _ {2} - 80 a _ {4} + 18 a _ {6} = 1$$

Equate the coefficients of $x \sin (3x)$ on both sides of the equation:

$$- 108 a _ {3} - 80 a _ {5} = 0$$

Equate the coefficients of $x^{2} \sin(3x)$ on both sides of the equation:

$$-80a_6 = 1$$

Solve the system:

$$\begin{array}{l} a_1 = 0 \\ a_2 = -\frac{27}{1600} \\ a_3 = 0 \\ a_4 = -\frac{251}{64000} \\ a_5 = 0 \\ a_6 = -\frac{1}{80} \\ \end{array}$$

Substitute $a_1, a_2, a_3, a_4, a_5$, and $a_6$ into

$$\begin{array}{l} y_p(x) = a_1 \cos(3x) + a_2 x \cos(3x) + a_3 x^2 \cos(3x) + \\ a_4 \sin(3x) + a_5 x \sin(3x) + a_6 x^2 \sin(3x) \\ \end{array}$$ $$y_p(x) = -\frac{27x \cos(3x)}{1600} - \frac{251 \sin(3x)}{64000} - \frac{1}{80} x^2 \sin(3x)$$

The general solution is:

$$\begin{array}{l} y(x) = y_c(x) + y_p(x) = \\ -\frac{27x \cos(3x)}{1600} - \frac{251 \sin(3x)}{64000} - \frac{1}{80} x^2 \sin(3x) + c_1 \cos\left(\frac{x}{3}\right) + c_2 \sin\left(\frac{x}{3}\right) \\ \end{array}$$