Toronto Math Forum

Transform the given system into a single equation of second order and find the solution $(x_1(t), x_2(t))$, satisfying initial conditions

$$\left\{\begin{aligned} & x'_1= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\ &x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2 \end{aligned}\right.$$


a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=\frac{1}{2}x’_1+\frac{1}{4}x_1\tag{1}$$

Plugging $x_2$ into the second equation gives

$$\left(\frac{1}{2}x’_1+\frac{1}{4}x_1\right)’=-2x_1-\frac{1}{2}\left(\frac{1}{2}x’_1+\frac{1}{4}x_1\tag{1}\right)$$
$$x''_1+x'_1+\frac{17}{4}x_1=0$$

b. Find $x_1$ and $x_2$ that also satisfy the given initial conditions.

The characteristic equation is
$$r^2+r+\frac{17}{4}=0$$
$$r_1,2=\frac{1}{2}\pm 2i.$$

The general solution is
$$x_1(t)=e^{-\frac{1}{2}t}\left(c_1\cos{2t}+c_2\sin{2t}\right).$$

It follows that the solution for $x_2$ is

$$x_2(t)=-\frac{1}{2}e^{-\frac{1}{2}t}(c_1\left(-\frac{1}{4}\cos{2t}-2\sin{2t}\right)+c_2\left(2\cos{2t}-\frac{1}{4}\sin{2t}\right))$$

Satisfying the given initial conditions  $x_1(0)=-2$ and $x_2(0)=2$, we have

$$\left\{\begin{aligned} &c_1=-2\\ &c_2=2 \end{aligned}\right.$$

The solutions that satisfy the given initial conditions are

$$x_1(t)=e^{-\frac{1}{2}t}(2\sin{2t}-2\cos{2t})$$
$$x_2(t)=e^{-\frac{1}{2}t}(2\cos{2t}+2\sin{2t}).$$

Transform the given system into a single equation of second order and find the solution (x_1(t), x_2(t)), satisfying initial conditions
$$\left\{\begin{aligned} &x'_1= 2x_2, &&x_1(0) = 3,\\ &x'_2= -2x_1, &&x_2(0) = 4. \end{aligned}\right.$$


a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=\frac{1}{2}x'_1$$

Plugging $x_2$ into the second equation gives

$$\left(\frac{1}{2}x’_1\right)’=-2x_1$$
$$\frac{1}{2}x’’_1+2x’_1=0$$
$$x’’_1+4x’_1=0$$

b. Find $x_1$ and $x_2$ that also satisfy the given initial conditions.

The characteristic equation is
$$\frac{1}{2}r^2+2=0$$
$$r^2+4=0$$
$$ r_1,2=\pm 2i$$

The general solution is
$$x_1(t)=c_1\cos{2t}+c_2\sin{2t}.$$

It follows that the solution for $x_2$ is

$$x_2(t)=\frac{1}{2}(-2c_1\sin{2t}+2c_2\cos{2t})=-c_1\sin{2t}+c_2 \cos{2t}$$

Satisfying the given initial conditions  $x_1(0)=3$ and $x_2(0)=4$, we have

$$\left\{\begin{aligned} &c_1=3\\ &c_2=4\end{aligned}\right.$$


The solutions that satisfy the given initial conditions are

$$x_1(t)=3\cos{2t}+4\sin{2t}$$
$$x_2(t)=-3\sin{2t}+4\cos{2t}.$$

Find the general solution of the given differential equation.

$$4y'' + y = 2 \sec\left(\frac{t}{2}\right)\tag{1},\qquad -\pi  < t < \pi.$$

Dividing $(1)$ by $4$, we have

$$y''+\frac{1}{4}y=\frac{1}{2}\sec{\frac{t}{2}},$$


where $p(t)=0$, $q(t)=\frac{1}{4}$, and $g(t)=\frac{1}{2}\sec{\frac{t}{2}}$ are continuous on $(-\pi, \pi)$.

The corresponding homogeneous equation is
$$4y''+y=0,$$

with characteristic equation
$$4r^2+1=0 \Rightarrow r_{1,2}=\pm\frac{i}{2}.$$

The homogeneous solution is $y_h(t)=c_1\cos\frac{t}{2}+c_2\sin\frac{t}{2}$


Computing the Wronskian,


$$W\left(\cos{\frac{t}{2}}, \sin{\frac{t}{2}}\right)(t)=\begin{array}{|c c|}\cos{\frac{t}{2}}& \sin{\frac{t}{2}} \\ -\frac{1}{2}\sin{\frac{t}{2}} &\frac{1}{2}\cos{\frac{t}{2}}\end{array}=\frac{1}{2}\ne0,$$

we verify that $y_1(t)=\cos{\frac{t}{2}}$ and $y_2(t)=\sin{\frac{t}{2}}$ form a fundamental set of solutions.


Calculating the parameters $u_1$ and $u_2$, we have

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\tan{\frac{t}{2}}}dt=2\ln\left(\cos{\frac{t}{2}}\right)+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{1}dt=t+c_2.$$

It follows that the general solution is

$$y(t)=\cos{\frac{t}{2}}\left(2\ln{\left(\cos{\frac{t}{2}}\right)}+c_1\right)+\sin{\frac{t}{2}}(t+c_2)=2\cos{\frac{t}{2}}\ln{\left(\cos{\frac{t}{2}}\right)}+t\sin{\frac{t}{2}}+c_1\cos{\frac{t}{2}}+c_2\sin{\frac{t}{2}}.$$

Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.

$$\begin{gather*} t^2y''-2y=3t^2-1,\qquad t>0\tag{1};\\  y_1(t)=t^2,\quad  y_2(t)=t^{-1}. \end{gather*}$$

1. Verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ satisfy the corresponding homogeneous equation,

$$t^{2}y''-2y=0.\tag{2}$$

$$\cases{y_1(t)=t^{2}\\y'_1(t)=2t\\y''_1(t)=2}$$

Plugging $y_1$, $y'_1$, and $y''_1$ into $(2)$,

$$t^{2}(2)-2(t^{2})=0 \Rightarrow 2t^{2}-2t^{2}=0$$

$y_1$ satisfies $(2)$ $\Rightarrow$ $y_1$ is a solution to the corresponding homogeneous equation.


Similarily,

$$\cases{y_2(t)=t^{-1}\\y'_2(t)=-t^{-2}\\y''_2(t)=2t^{-3}}$$

Plugging $y_2$, $y'_2$, and $y''_2$ into $(2)$,

$$t^{2}\left(\frac{2}{t^{3}}\right)-2\left(\frac{1}{t}\right)=0 \Rightarrow \frac{2}{t}-\frac{2}{t}=0$$

$y_2$ satisfies $(2)$ $\Rightarrow$ $y_2$ is a solution to the corresponding homogeneous equation.


2. Find a particular solution of the given nonhomogeneous equation,

$$t^{2}y''-2y=3t^{2}-1, t>0.\tag{1}$$

Dividing $(1)$ by $t^{2}$, we have

$$y''-\frac{2}{t^{2}}y=3-\frac{1}{t^{2}}$$

Note that $p(t)=0$, $q(t)=-\frac{2}{t^{2}}$, and $g(t)=3-\frac{1}{t^{2}}$ are continuous on $(0, +\infty)$

Computing the Wronskian,

$$W(t^{2}, t^{-1})(t)=\begin{array}{|c c|}t^{2}& t^{-1} \\ 2t & -t^{-2}\end{array}=-1-2=-3\ne0,$$

we verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ form a fundamental set of solutions $\forall t>0$

Calculating the parameters $u_1$ and $u_2$, we get

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\frac{t^{-1}(3-t^{-2})}{-3}}dt=\frac{1}{3}\int{\frac{3}{t}-\frac{1}{t^{3}}}dt=\ln{t}+\frac{1}{6t^{2}}+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{\frac{t^{2}(3-t^{-2})}{-3}}dt=-\frac{1}{3}\int{3t^{2}-1}dt=-\frac{t^{3}}{3}+\frac{t}{3}+c_2$$

The particular solution is

$$y_p(t)=t^{2}\left(\ln{t}+\frac{1}{6t^{2}}+c_1\right)+\frac{1}{t}\left(-\frac{t^{3}}{3}+\frac{t}{3}+c_2\right)=t^{2}\ln{t}+\frac{1}{6}+c_1t^{2}-\frac{t^{2}}{3}+\frac{1}{3}+\frac{c_2}{t}=t^{2}\ln{t}+\frac{1}{2}+c_1t^{2}+\frac{c_2}{t}$$

$$y_p(t)=t^{2}\ln{t}+\frac{1}{2}.$$

(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE

$$-y''(e^{x}+e^{-x}+2)+y'(e^{x}-e^{-x})+2y=0.$$

Dividing but sides by $-(e^{x}+e^{-x}+2)$, we get

$$L[y]=y''-y'\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}-y\frac{2}{e^{x}+e^{-x}+2}=0,$$

where $p(x)=-\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}$, and $q(t)=-\frac{2}{e^{x}+e^{-x}+2}$.

By Abel's Theorem,

$$\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}dx)\\&=c (e^x+e^{-x}+2).\end{align}$$

Let $c=1 \Rightarrow W(y_1,y_2)(x)=e^x+e^{-x}+2$.

b) Check that $y_1(x)=e^{x}+1$ is a solution and find another linearly independent solution.

Since $y_1(x)=e^{x}+1 \Rightarrow y_1 '(x)=e^{x}$, and $y_1 ''(x)=e^{x}$

Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have

$$\begin{align}-e^{x}(e^{x}+e^{-x}+2)+e^{x}(e^{x}-e^{-x})+2(e^{x}+1)&=0\\-2e^x-e^{2x}-2+2e^{x}+e^{2x}+2&=0\end{align}$$

$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.

Given $y_1(x)$, we can find another linearly independent solution.

We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=(e^x+1)y_2 '-y_2 e^x$$

Equating the two expressions for the Wronskian, we get

$$(e^x+1)y_2 '-y_2 e^x=e^x+e^{-x}+2$$

Dividing both sides by $e^x+1$, and multiplying by integrating factor $\mu=\frac{1}{e^x+1}$,


$$\frac{1}{e^x+1}y_2=\int{\frac{e^x+e^{-x}+2}{e^{2x}+2e^x+1}}dx+C$$
$$\frac{1}{e^x+1}y_2=-\frac{1}{e^x}$$
$$y_2=-\frac{e^x+1}{e^x}$$
$$y_2=-1-\frac{1}{e^{x}}$$

c) Write the general solution. Find solution such that $y(0)=0$, $y'(0)=2$

The general solution to the ODE is

$$y(x)=c_1 (e^x+1)+c_2(-1-\frac{1}{e^{x}}).$$

$\Rightarrow y'(x)=c_1e^x+c_2e^{-x}$

$$\cases{c_1-c_2=0\\c_1+c_2=2} \Rightarrow \cases{c_1=1\\c_2=1}$$

Thus, the solution that satisfies $y(0)=1$, $y'(0)=2$ is

$$y(x)=e^x-\frac{1}{e^{x}}.$$

(Pengyun's solution with corrected derivative of $xe^{x}$)

Find the Wronskian of the given pair of functions: $x$ and $xe^{x}$


$$W(x, xe^x) = \left|\begin{matrix}x & xe^{x} \\ x' & (xe^{x})'\end{matrix}\right|= \left|\begin{matrix}x & xe^{x} \\ 1 & xe^{x}+e^{x}\end{matrix}\right| = x^{2}e^{x}+xe^{x}-xe^{x}=x^{2}e^{x}.$$

Find general and special solutions to $$y= 2xy'-3(y')^2.$$

Substituting $y'=p$, we get an equation of the form $y=2xp–3{p^2}$

Differentiating both sides, we get

$dy=2xdp+2pdx-6pdp$

$dy=pdx$   $\Rightarrow$   $pdx=2xdp+2pdx-6pdp$    $\Rightarrow$    $–pdx=2xdp–6pdp$

Dividing by $p$, we get $-dx={2x \over p}dp-6dp$   $\Rightarrow$   $ {dx \over dp} + {2 \over p}x-6=0  \tag{1}$

We get a linear differential equation. To solve $(1)$, we use an integrating factor given by $\mu=e^{\int{2 \over p}dp}=e^{2\ln|p|}=p^2$

Rewrite $(1)$ as ${2 \over p}x+{dx \over dp}=6$ and multiply by the integrating factor, $\mu(p)=p^2$ to get,

$$p^2{2 \over p}x+p^2{dx \over dp}=p^2 6$$
$${d \over dp}(p^2 x)=p^2 6$$
$$x={{2p^3+C} \over p^2}$$


Thus, the general solution to (1) is
$$x={{2p + {C \over p^2}}}\tag{2}$$

Substituting $(2)$ into the Lagrange Equation, we get


$$y={2\left({2p+\frac{C}{{{p^2}}}}\right)p–3p^2}=p^2+ \frac{2C}p$$

Thus,

\begin{equation} \left\{\begin{aligned} &x={{2p + {C \over p^2}}}\\ &y=p^2+ \frac{2C}p \end{aligned}\right. \tag{3}\end{equation}


gives us a solution in the parametric form.


The Lagrange equation, $y= 2xy'-3(y')^2$, can also have a special solution (or solutions)

$\varphi (p)-p=0$

$2p–p=0$   $\Rightarrow$    $p(2-p)=0$    $\Rightarrow$    $p_1=0$ and $p_2=2$


Thus, the special solutions are
$$y=x\varphi (0)+\psi (0)=x \cdot 0+0=0.$$