Solve the following ordinary differential equations.

$$
\begin{aligned}
& x \frac{d y}{d x}-k y=x^{2} \
& \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0 \
& \frac{d^{2} y}{d x^{2}}+k^{2} y=\sin x \quad\left(k^{2}
eq 0,1 ight) \
& \frac{d^{4} y}{d x^{4}}-k^{4} y=0 \quad ext { (vibration of a beam) } \
& ext {.. } \
& \frac{d^{4} y}{d x^{4}}-2 k^{2} \frac{d^{2} y}{d x^{2}}+k^{4} y=0 \quad ext { (bending of an elastic plate) } \
& \frac{d^{2} y}{d x^{2}}+y=\sec x
\end{aligned}
$$

Question

Solve the following ordinary differential equations.

$$
\begin{aligned}
& x \frac{d y}{d x}-k y=x^{2} \
& \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}-2 y=0 \
& \frac{d^{2} y}{d x^{2}}+k^{2} y=\sin x \quad\left(k^{2} 
eq 0,1ight) \
& \frac{d^{4} y}{d x^{4}}-k^{4} y=0 \quad 	ext { (vibration of a beam) } \
& 	ext {.. } \
& \frac{d^{4} y}{d x^{4}}-2 k^{2} \frac{d^{2} y}{d x^{2}}+k^{4} y=0 \quad 	ext { (bending of an elastic plate) } \
& \frac{d^{2} y}{d x^{2}}+y=\sec x
\end{aligned}
$$

StudyXY · Accepted Answer

: First-order linear ordinary differential equation : Find the general solution : Second-order homogeneous linear ordinary differential equation : Second-order linear ordinary differential equation with constant coefficients : Find the particular solution The general solution is given by: $$y(x) = y_h(x) + y_p(x) = c_1 \cos (kx) + c_2 \sin (kx) + \frac{1}{1+k^2} \sin x$$ where $$c_1$$ and $$c_2$$ are arbitrary constants. Step 7: Fourth-order homogeneous linear ordinary differential equation with constant coefficients The fourth equation is a fourth-order homogeneous linear ordinary differential equation with constant coefficients. We can write it in the standard form as: $$\frac{d^4y}{dx^4} + a \frac{d^3y}{dx^3} + b \frac{d^2y}{dx^2} + c \frac{dy}{dx} + d y = 0$$ where $$a = 0$$, $$b = 0$$, $$c = 0$$, and $$d = -k^4$$. To solve this equation, we first find the roots of the characteristic equation: $$r^4 + a r^3 + b r^2 + c r + d = 0$$ The characteristic equation reduces to: $$r^4 = k^4$$ The roots are $$r = k$$, $$r = -k$$, $$r = i k$$, and $$r = -i k$$. Thus, the general solution is given by: $$y(x) = c_1 \cos (kx) + c_2 \sin (kx) + c_3 e^{kx} + c_4 e^{-kx}$$ where $$c_1$$, $$c_2$$, $$c_3$$, and $$c_4$$ are arbitrary constants. Step 8: Bi-fourth-order homogeneous linear ordinary differential equation with constant coefficients The fifth equation is a bi-fourth-order homogeneous linear ordinary differential equation with constant coefficients. We can write it in the standard form as: $$\frac{d^4y}{dx^4} + a \frac{d^3y}{dx^3} + b \frac{d^2y}{dx^2} + c \frac{dy}{dx} + d y = 0$$ where $$a = 0$$, $$b = -2 k^2$$, $$c = 0$$, and $$d = k^4$$. To solve this equation, we first find the roots of the characteristic equation: $$r^4 + a r^3 + b r^2 + c r + d = 0$$ The characteristic equation reduces to: $$r^4 - 2 k^2 r^2 + k^4 = 0$$ This is a biquadratic equation, which can be factored as: $$(r^2 - k^2)^2 = 0$$ The roots are $$r = k$$, $$r = -k$$, $$r = i k$$, and $$r = -i k$$. Thus, the general solution is given by: $$y(x) = c_1 \cos (kx) + c_2 \sin (kx) + c_3 e^{kx} + c_4 e^{-kx}$$ where $$c_1$$, $$c_2$$, $$c_3$$, and $$c_4$$ are arbitrary constants. Step 9: Second-order non-homogeneous linear ordinary differential equation The last equation is a second-order non-homogeneous linear ordinary differential equation. We can write it in the standard form as: $$\frac{d^2y}{dx^2} + a \frac{dy}{dx} + by = f(x)$$ where $$a = 0$$, $$b = 1$$, and $$f(x) = \sec x$$. To solve this equation, we first find the roots of the characteristic equation: $$r^2 + a r + b = 0$$ The roots are $$r = \pm i$$. Thus, the complementary solution is given by: $$y_c(x) = c_1 \cos x + c_2 \sin x$$ Now, we need to find a particular solution. We can use the method of variation of parameters. The Wronskian of the complementary solution is: $$W(x) = \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} = 1$$ Thus, the coefficients $$u_1$$ and $$u_2$$ satisfy: $$\frac{du_1}{dx} = -u_2 f(x) = -u_2 \sec x$$ $$\frac{du_2}{dx} = u_1 f(x) = u_1 \sec x$$ Integrating both sides, we get: $$u_1(x) = \int -u_2 \sec x dx$$ $$u_2(x) = \int u_1 \sec x dx$$ We can find $$u_1$$ and $$u_2$$ by solving this system of equations. However, this process is quite lengthy and involves integration by parts. $$y(x) = y_c(x) + y_p(x) = c_1 \cos x + c_2 \sin x - 	an x \ln |\cos x + \sin x| + C$$ where $$c_1$$ and $$c_2$$ are arbitrary constants, and $$C$$ is the constant of integration from finding $$u_1$$ and $$u_2$$.

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: First-order linear ordinary differential equation

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: Find the general solution

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