7.4 Series solutions of differential equations  (Page 2/2)

$y\text{″}+6y^{\prime }=0$

$5y\text{″}+y^{\prime }=0$

$y\text{″}+25y=0$

$y\text{″}-y=0$

$2y^{\prime }+y=0$

$y^{\prime }-2xy=0$

$\left(x-7\right)y^{\prime }+2y=0$

$y\text{″}-x{y}^{\prime }-y=0$

$\left(1+x^2\right)y\text{″}-4x{y}^{\prime }+6y=0$

$x^{2}y\text{″}-x{y}^{\prime }-3y=0$

$y\text{″}-8y^{\prime }=0,y\left(0\right)=\mathrm{-2},y^{\prime }\left(0\right)=10$

$y\text{″}-2xy=0,y\left(0\right)=1,y^{\prime }\left(0\right)=\mathrm{-3}$

The differential equation $x^{2}y\text{″}+x{y}^{\prime }+\left(x^2-1\right)y=0$ is a Bessel equation of order 1. Use a power series of the form $y={\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ to find the solution.

If $y$ and $z$ are both solutions to $y\text{″}+2y^{\prime }+y=0,$ then $y+z$ is also a solution.

The following system of algebraic equations has a unique solution:

$\begin{array}{c}6z_1+3z_2=8\hfill \\ 4z_1+2z_2=4.\hfill \end{array}$

$y=e^x\text{cos}(3x)+e^x\text{sin}(2x)$ is a solution to the second-order differential equation $y\text{″}+2y\prime +10=0.$

To find the particular solution to a second-order differential equation, you need one initial condition.

$y\text{″}-2y=0$

$y\text{″}-3y+2y=\text{cos}\left(t\right)$

${\left(\frac{dy}{dt}\right)}^2+y{y}^{\prime }=1$

$\frac{d^{2}y}{d{t}^2}+t\frac{dy}{dt}+{\text{sin}}^2\left(t\right)y=e^t$

$y\text{″}+9y=0$

$y\text{″}+2y^{\prime }+y=0$

$y\text{″}-2y^{\prime }+10y=4x$

$y\text{″}=\text{cos}\left(x\right)+2y^{\prime }+y$

$y\text{″}+5y+y=x+e^{2x}$

$y\text{″}=3y^{\prime }+x{e}^{\text{−}x}$

$y\text{″}-x^2=\mathrm{-3}{y}^{\prime }-\frac{9}{4}y+3x$

$y\text{″}=2\text{cos}x+y^{\prime }-y$

$y\text{″}+4y^{\prime }+6y=0,$ $y\left(0\right)=0,$ $y^{\prime }\left(0\right)=\sqrt{2}$

$y\text{″}=3y-\text{cos}\left(x\right),$ $y\left(0\right)=\frac{9}{4},$ $y^{\prime }\left(0\right)=0$

$4y^{\prime }=\mathrm{-6}y+2y\text{″},$ $y\left(0\right)=0,$ $y\left(1\right)=1$

$y\text{″}=3x-y-y^{\prime },$ $y\left(0\right)=\mathrm{-3},$ $y\left(1\right)=0$

The motion of a swinging pendulum for small angles $\theta$ can be approximated by $\frac{d^2\theta }{d{t}^2}+\frac{g}{L}\theta =0,$ where $\theta$ is the angle the pendulum makes with respect to a vertical line, g is the acceleration resulting from gravity, and L is the length of the pendulum. Find the equation describing the angle of the pendulum at time $t,$ assuming an initial displacement of ${\theta }_0$ and an initial velocity of zero.

Find the general solution to this equation ( Hint: call ${\omega }_0=\sqrt{b\text{/}a}$ ).

Assuming the system starts from rest, show that the particular solution can be written as $y=\frac{2}{a\left({\omega }_0{}^2-{\omega }^2\right)}\text{sin}\left(\frac{{\omega }_0-\omega t}{2}\right)\text{sin}\left(\frac{{\omega }_0+\omega t}{2}\right).$

[T] Using your solutions derived earlier, plot the solution to the system $2y\text{″}+9y=\text{cos}(2t)$ over the interval $t=[\mathrm{-50},50].$ Find, analytically, the period of the fast and slow amplitudes.

An opera singer is attempting to shatter a glass by singing a particular note. The vibrations of the glass can be modeled by $y\text{″}+ay=\text{cos}(bt),$ where $y\text{″}+ay=0$ represents the natural frequency of the glass and the singer is forcing the vibrations at $\text{cos}(bt\text{).}$ For what value $b$ would the singer be able to break that glass? ( Note : in order for the glass to break, the oscillations would need to get higher and higher.)