Differential equations

$x^2=x$, what will we result in this equation? A set of numbers.

$y^{'}=y$, what will we result in this equation? A set of functions.

What are differential equations?

A differential equation is an equation that involves one or more derivatives of an unknown function, which one will have to find because it satisfies the differential equation on some open interval.

What are ordinary differential equations (ODE)?

When your equation involves a function with respect to only one variable $x$ in the $Domain$, i.e. $y=f(x)$. Thus,

N-th order equation and his grade

The order equation is the order of his highest derivative.

Grade equation is the algebraic grade of the grade of his highest derivatives.

Applications.

Newton's Law of Cooling

Newton's Second Law of Motion

What are partial differential equations (PDE)?

When your equation involves a function for a space $x$.

Applications.

Laplace's equation

Heat equation

Graphical and Numerical Methods

Graphical methods or direction fields

The direction field gives a picture of the first-order equation and its integral curves give a picture of the solutions.

Sketch $y'=f(x,y)$ with representative set of points in the plane (x,y).

Methods

The simplest of all differential equations is

and we solve it by

Cauchy problems or initial value problem

Sometimes we are interested in a particular solution given an initial condition, that is, our solution only satisfies this condition.

Example.

Existence and Uniqueness Theorem or Picard–Lindelöf theorem.

If $f(x,y)$ and $\dfrac{df(x,y)}{dy}$ are continuous functions on a closed rectangle R,

then by each point $(x_0,y_0)$ in the interior of R there passes a unique integral curve of the equation $\dfrac{dy}{dx}=f(x,y)$

Application.

Worked examples

• Find m such that $y=e^{mx}$ in $y^{''}-6y^{'}+8y=0$ be true.

  Substitute and calculate his derivatives.

  $m^2e^{mx}-6me^{mx}+8e^{mx}=0$

  $m^2-6m+8=0$

  $m_1=4$ $m_2=2$

• Resolve $y^{'''}=sin(x)$ with $y^{''}(0)=y^{'}(0)=y(0)=0$

  $y^{'''}=sin(x)$

  $y^{''}=-cos(x)+c_1$

  $y^{'}=-sin(x)+c_1x+c_2$

  $y=cos(x)+c_1\dfrac{x^2}{2}+c_2x+c_3$

  But

  $y^{''}(0)=-cos(0)+c_1=0\rarr c_1=1$

  $y^{'}(0)=c_2=0$

  $y(0)=1+c_3=0\rarr c_3=-1$

  Therefore, the particular solution is

  $y=cos(x)+\dfrac{x^2}{2}-1$

• Resolve general Newton's Law of Cooling

Separable equations

• Proof.

  $$\int h(y)\dfrac{dy}{dx}dx=\int g(x)dx\\ \dfrac{dH}{dy}=h(y)\\ \dfrac{dH}{dx}=\dfrac{dH}{dy}\dfrac{dy}{dx}=h(y)\dfrac{dy}{dx}\\ \int \dfrac{dH}{dx}dx=\int g(x)dx\\ H(y)=\int g(x)dx\\ \int h(y)dy = \int g(x)dx$$

Worked examples

• $1+e^{4\pi x}y'=0$

  $$y'=\dfrac{-1}{e^{4\pi x}}\\ \int \dfrac{dy}{dx}dx=\int\dfrac{-1}{e^{4\pi x}}dx\\ y=\dfrac{1}{4\pi}e^{-4\pi x}+C$$

• $\dfrac{1}{y'}=\dfrac{x^2y^{\pi-1}}{1+x}$

  $$y'=\dfrac{1+x}{x^2y^{\pi-1}}\\ y^{\pi-1}y'=\dfrac{1+x}{x^2}\\ \int y^{\pi-1}y'dy=\int1/x^2dx+\int\dfrac{1}{x}dx\\ y^\pi=-\dfrac{\pi}{x}+\pi ln|x|+c$$

• $\dfrac{1}{2}(x+1)y'=x\sqrt{y}$

  $$\dfrac{1}{\sqrt y}y'=\dfrac{2x}{x+1}\\ \int\dfrac{1}{\sqrt y}dy=\int \dfrac{2x}{x+1}dx\\ 2\sqrt{y}=2x-2ln|x+1|+C_0\\ \sqrt{y}=x-ln|x+1|+C_1\\ y=(x-ln|x+1|+C_1)^2$$

• $xy'=180y$

  $$\dfrac{1}{y}y'=\dfrac{180}{x}\\ \int\dfrac{1}{y}dy=\int\dfrac{180}{x}dx\\ ln|y|=180ln|x|+ln|c_1|\\ ln|y|=ln|c_1x^{180}|\\ y=Cx^{180}$$

• $(e^{-y}+3)cos(x)dx=(9-3sin(x))dy$ such that $y(0)=0$

  $$\dfrac{3}{e^{-y}+3}y'=\dfrac{cos(x)}{3-sin(x)}\\ \int\dfrac{3}{e^{-y}+3}dy=\int\dfrac{cos(x)}{3-sin(x)}dx\\ ln|1+3e^y|=ln|c_1|-ln|3-sin(x)|\\ 1+3e^y=\dfrac{c_1}{3-sin(x)}\\ c_1=(1+3e^y)(3-sin(x))\\ \text{But y(0)=0}\\ c_1=(1+3e^0)(3-sin(0))\\ c_1=12\\ (1+3e^y)(3-sin(x))=12$$

Radiocarbon dating, developed by the American chemist Willard Libby in 1949 and for which he won the Nobel Prize in Chemistry in 1960, is a method that uses the carbon-14 isotope to estimate the age of organic materials up to around 50,000 years. The method is based on the fact that the decay rate of carbon-14 is directly proportional to the number of atoms that the material to be evaluated has over time $t$. Added to this, the half-life (or half-life) of this isotope is known to be approximately 5,730 years. Therefore, we can use differential equations to model such radioactive decay and thus estimate the age of a certain material. According to A. J. Timothy Jull and other authors, in their research paper Radiocarbon Dating of the Scrolls and Linen Fragments from the Judean Desert, '' in March 1994, radiocarbon dating was used to determining the age of the Book of the Prophet Isaiah. The researchers took a sample of the manuscript and determined that it contained approximately $75\%$ the initial amount of carbon-14. By GalileoX CMath002X Introducción a las ecuaciones diferenciales.

• a) Determine the amount of carbon-14 present in the manuscript sample as a function of time. Assume $k$ that it is the constant of proportionality of the model and that $N_0$ it is the initial amount of carbon-14.

  $$N'\propto N \\N'-kN=0,k<0 \text{ Since this is a decreasing problem}\\ N(0)=N_0\\ N(t)=N_0e^{-P(t)},P(t)=\int-kdt\\ N(t)=N_0e^{-kt}$$

• Determine the constant of proportionality k of the previous model.

  $$N(5730)=N_0/2\\ N(0)=N_0=0.75\\ N(t)=N_05e^{kt},k<0\\ N(5730)=N_0e^{k5730}=N_0/0.5\\ 5730k=ln(2)\\ k=-\dfrac{ln(2)}{5730}$$

• Estimate the date the manuscript was written.

  $$0.75N_0=N_0e^{kt^*}\\ t^*=\dfrac{ln(0.75)}{k}=2378.16\\ 1994-2378=-384\rarr \text{384 AC}$$

An industrial process requires raising the temperature of a certain material to $100°C$. After the molding process has been carried out, the part is placed in a room with a controlled temperature of $20°C$. An operator determines that after 20 minutes the temperature of the part is $40°C$. For the piece to be considered ready to go to the next production phase, it must reach the $25°C$.

• Determine the time (in minutes) required for the part to be ready to go to the next phase.

  $$\dfrac{dT}{dt}=k(T-20)\\ T(0)=100\\ T(20)=40\\ T(t^*)=25\\ \int\dfrac{1}{T-20}dT=k\int dt\\ ln|T-20|=kt+C\\ T-20=Ce^{kt}\\ T(t)=Ce^{kt}+20\\ T(0)=C+20=100,C=80\\ T(20)=80e^{k20}+20=40,k=ln(\dfrac{1}{4})/20=-0.06931471805\\ T(t)=80e^{kt}+20\\ T(t^*)=80e^{kt^*}+20=25\\ 80e^{kt^*}=5\\ e^{kt^*}=\dfrac{1}{16}\\ kt^*=ln(\dfrac{1}{16})\\ t^*=20\dfrac{ln(\dfrac{1}{16})}{ln(\dfrac{1}{4})}=\text{40 min}$$

Suppose that velocity of a sphere decreases directly proportional to its surface area. The sphere starts with radius 1. After one minute, the measured radius is half the initial radius.

• Calculate time (in minutes) to disintegrate completely.

  $$\dfrac{dV}{dt}\propto A(r)\\ \dfrac{dV}{dt}=kA(r)\\ \dfrac{dV}{dr}\dfrac{dr}{dt}=kA(r)\\ (4\pi r^2)\dfrac{dr}{dt}=k(4\pi r^2)\\ \dfrac{dr}{dt}=k\\ \int dr=\int kdt\\ r(t)=kt+C\\ \text{But }r(0)=k\times0+C=1,C=1\\ r(1)=k\times1+1=\dfrac{1}{2},k=-1/2\\ \text{Therefore, } r(t^*)=-\dfrac{1}{2}t^*+1=0, \text{ when } t^*=\text{2 min}$$

First-order equations

Linear equations

I guess it $y_n(x)$is the general solution from an homogeneous equation associated and $y_p(x)$ is a particular solution from lineal linear equation. So,

$y_n'=f_1y_n$ and $y_p'=f_1y_p+f_2$

$y'_n+y_p'=f_1y_n+f_1y_p+f_2$

Since $\dfrac{d}{dx}[f_1(x)+f_2(x)]=\dfrac{df_1}{dx}+\dfrac{df_2}{dx}$,

$(y_n+y_p)'=f_1(y_n+y_p)+f_2$

As $y=y_n+y_p$

$y'=f_1y+f_2$

Solution for Homogeneous equation associated

As $y_h'=f_1y_h$

$\dfrac{y_h'}{y_h}=f_1(x)\rarr ln|y|=ln|c|=\int f_1(x)dx$

$y_h(x)=ce^{\int f_1(x)dx}$

Solution algorithm for particular solution $y_p$ by Lagrange method

$y(x)=z(x)y_h(x)$

$z(x)=\int\dfrac{f_2(x)}{y_h(x)}dx$

• Proof

  $$y=zy_h\\ y'=z'y_h+zy_h'\\ \text{Substitute in } y'=f_1y+f_2\\ z'y_h+zy_h'=f_1(z_yy_h)+f_2\\ \text{But } y_h'=f_1y_h\\ z'y_h+zf_1y_h=f_1zy_h+f_2\\ z'y_h=f_2\\ z'=\dfrac{f_2}{y_h}\\ z(x)=\int \dfrac{f_2(x)}{y_h(x)}dx$$

If $y'+f_1(x)y=f_2(x)$,

then $y(x)=ce^{\int f_1(x)dx}(\int\dfrac{f_2(x)}{ce^{\int f_1(x)dx}}dx)$

Solution algorithm for particular solution $y_p$ by Method of undetermined coefficients

Worked examples

• $y'=-7y+17t, y(0)=-9/7$

  $$\text{Problem: }y'+7y=17t,y(0)=-9/7\\ \mu=e^{\int7dt}=e^{7t}\\ y(t)=\dfrac{1}{e^{7t}}\int17te^{7t}dt\\ y(t)=\dfrac{1}{e^{7t}}(\dfrac{17te^{7t}}{7}-\dfrac{17e^{7t}}{49}+c)=\dfrac{17t}{7}-\dfrac{17}{7}+\dfrac{c}{e^{7t}}\\ y(0)=-\dfrac{17}{7}+c=-\dfrac{9}{7}\rarr c=\dfrac{8}{7}\\ \text{Particular solution: } y(t)=\dfrac{17t}{7}-\dfrac{17}{7}+\dfrac{8}{7e^{7t}}$$

• Mediante la EDO $S'(t)=rS(t)+k(t)$ puede modelarse el valor de una inversión $S$ respecto al tiempo $t$ donde $r$ es la tasa de interés y $k(t)$ es un monto de dinero que se deposita al tiempo $t$. Resuelve el PVI definido por la EDO, la condición inicial $S(0)=0$, una tasa de interés $r=0.06$ y $k(t)=1000e^{rt}$.

  $$Problem: S'(t)-rS(t)=k(t),r\in\mathbb{R}\\ \mu=e^{\int -r dt}=e^{-rt}\\ S(t)=\dfrac{1}{e^{-rt}}\int 1000e^{rt}e^{-rt}dt=\dfrac{1}{e^{-rt}}(1000t+c)\\ S(0)=0+c=0\implies c=0\\ \text{Particular solution: } S(t)=\dfrac{1000t}{e^{-0.06t}}$$

• Un tanque tiene 20 kg de sal disueltos en 100 l de agua. Al tanque entra una mezcla de agua con sal a una tasa de 12 l/min con una concentración de 3/1000 kg/l de sal. Además, el contenido del tanque sale a una rapidez de 10 l/min. ¿Cuánta sal habrá en el tanque a los 150 min?

  $$Problem:salt'(t)=x'(t)=r_{in}c_{in}-r_{out}c_{out}(t),x(0)=20,V(0)=100\\ x'(t)=r_{in}c_{in}-r_{out}\dfrac{x(t)}{V(t)}\\ x'(t)+r_{out}\dfrac{x(t)}{V(t)}=r_{in}c_{in},V(t)=(r_{in}-r_{out})t+c_1\\ \mu=e^{r_{out}{\int \dfrac{1}{(r_{in}-r_{out})t+c}dt}}=e^{ln((r_{in}-r_{out})t+c)^{r_{out}})}=((r_{in}-r_{out})t+c)^{r_{out}}\\ \mu =(V(t))^{r_{out}}\\ x(t)=\dfrac{1}{(V(t))^{r_{out}}}\int r_{in}c_{in} (V(t))^{r_{out}}dt\\ x(t)=\dfrac{1}{(V(t))^{r_{out}}}(r_{in}c_{in})\dfrac{(V(t))^{r_{out}+1}}{{r_{out}+1}}+\dfrac{c_2}{(V(t))^{r_{out}}}\\ x(t)=\dfrac{r_{in}c_{in}V(t)}{{r_{out}+1}}+\dfrac{c_2}{(V(t))^{r_{out}}}\\ \text{Substituting}\\ x(t)=\dfrac{12*0.003 ((12-10)t+100)}{{{10}+1}}+\dfrac{c_2}{((12-10)t+100)^{{10}}}\\ x(t)=\dfrac{0.036 (2t+100)}{{11}}+\dfrac{1.6727273e+21}{(2t+100)^{{10}}}\\ x(15)=?$$

• Un objeto de masa $m$ se lanza verticalmente hacia abajo con una velocidad inicial $v_0$ desde una altura $h$. Aplique la Segunda Ley de Newton para determinar una expresión que describa la velocidad $v$ y posición  del objeto como una función del tiempo $t$. Determine la ecuación que describe la velocidad del objeto en cualquier instante de tiempo $t$, tomando en cuenta que $x(0)=h$. Encuentre el tiempo que tarda el objeto en llegar al suelo si $v_0=0 \text{ } m/s$ y $h=5 \text{ m}$.

  

  Por la segunda ley de Netwon:

  $$\sum F=ma$$

  Para nuestro caso de estudio:

  $$m(-g)=ma$$

  Donde $g=9.8\text{ }m/s^2$ y por definicion $a=\dfrac{dV}{dt}$

  Asi,

  $$\dfrac{dV}{dt}=-g\\ \int \dfrac{dV}{dt}dt=\int(-g)dt\\ V(t)=-gt+C$$

  Pero, $V(0)=V_0=C$

  Asi,

  $$V(t)=-gt+V_0$$

  Por definicion de velocidad

  $$V(t)=\dfrac{dx}{dt}\\ \int\dfrac{dx}{dt}dt=-gt+V_0\\ x(t)=-\dfrac{gt^2}{2}+V_0t+C$$

  Sin embargo, $x(0)=h=C$, obteniendo

  $$x(t)=-\dfrac{gt^2}{2}+V_0t+h$$

  Para la situacion particular $v_0=0 \text{ } m/s$ y $h=5 \text{ m}$, sustituimos en la ecuacion

  $$x(t)=-4.905t^2+5$$

  El suelo significa $x(t_1)=0$, donde $t_1>0$ es el tiempo buscado:

  $$-4.905t_1^2+5=0\\ t_1=\dfrac{\sqrt{4\cdot4.905\cdot5}}{(9.81)}\text{ s}\\ t_1=1.00963755469 \text{ s}$$

• Determine la ecuación de la curva $C$, en el plano, cuya recta tangente $T$ en cualquier punto $P(x,y)$ verifica la propiedad que la suma de sus interceptos con los ejes coordenados es una constant $k$. La curva pasa por el punto $(1,1)$.

  $$T_1=a+b=k$$

  where

  $$y'=\dfrac{y}{x-a}\rarr a=\dfrac{xy'-y}{y'}\\ y'=\dfrac{y-b}{x}\rarr b=y-xy'\\$$

  Thus,

  $$(\dfrac{xy'-y}{y'})+(y-xy')=k\\ \text{Clairaut's equation: }y=y'x+\dfrac{ky'}{y'-1},y(1)=1\\ y=cx+f(c)\rarr y=cx+\dfrac{kc}{c-1}\\ c=\dfrac{1}{2}(-k\pm\sqrt{k-4}\sqrt{k}+2)$$

  https://www.geogebra.org/calculator/z2qbh2y3

Homogeneous Function

Where $f:V\rarr W$ is a homogeneous function between two vector spaces over a field F and degree k as an integer.

Homogeneous differential equation, k=0

Solution algorithm

Preconditions. Check if the equation is a homogeneous differential equation, k=0.

1. Define $z=\dfrac{y}{x}$, so $y=zx\rarr$$y'=z'x+z$

2. Substitute (2) into the equation (1).

   $y'=f(\dfrac{y}{x})$

   $z'x+z=f(z)$

   $$\int \dfrac{1}{f(z)-z} dz=ln|c_0x|\\ \int \dfrac{1}{\dfrac{z-1}{z+1}-z}dz=ln|c_0x|\\ \int \dfrac{1}{\dfrac{(z-1)-z(z+1)}{z+1}}dz=ln|c_0x|\\ \int \dfrac{z+1}{(z-1)-(z^2+z)}dz=ln|c_0x|\\ \int \dfrac{z+1}{-1-z^2} dz = ln|c_0x| \\ -\dfrac{1}{2}ln(z^2+1)-arctan(z)=ln|c_0x|\\ -\dfrac{1}{2}ln(\dfrac{y}{x}^2+1)-arctan(\dfrac{y}{x})=ln|c_0x|\\ ln|(x^2+y^2)\dfrac{1}{2}|+arctan(\dfrac{y}{x})=c$$

   $z'x=f(z)-z$

   $\dfrac{z'}{f(z)-z}=\dfrac{1}{x}$ (Separable Differential equation)

3. Solve separable differential equation.

4. Check singular solution if $z=c\rarr z'=0$ such that $f(z)-z=0,$ so $z=\dfrac{y}{x}$

Post-conditions. Re-substitute function and singular solutions.

Worked examples.

• Given the ordinary differential equation $y^k+yx^{k-1}-x^ky'=0$, $k\in \mathbb{N},k>1.$ Identify his class and solve it.

  $y^k+yx^{k-1}-x^ky'=0$

  $y'=\dfrac{y^k}{x^k}+\dfrac{y}{x}$

  Therefore, this ordinary differential equation is a homogeneous differential equation.

  $$\int \dfrac{1}{z^k}dz=\int \dfrac{1}{x}dx\\ \dfrac{1}{(1-k)z^{1-k}}=ln|x|+c_0\\ \dfrac{1}{(1-k)z^{1-k}}=ln|c_1x|\\ \dfrac{1}{z^{1-k}}=(1-k)ln|c_1x|\\ \dfrac{x^{1-k}}{y^{1-k}}=ln|c_1x|-ln|c_2x^k|\\ \dfrac{x^{1-k}}{y^{1-k}}=ln|\dfrac{c_1x}{c_2x^k}|\\ \dfrac{x^{1-k}}{y^{1-k}}=ln|c_3x^{1-k}|\\ y^{1-k}=\dfrac{x^{1-k}}{ln|c_3x^{1-k}|}$$

• Given the ordinary differential equation $y'=\dfrac{y-x}{y+x}$. Identify his class and solve it.

  $$y'=\dfrac{y-x}{y+x}\dfrac{(\dfrac{1}{x})}{(\dfrac{1}{x})}\\ y'=\dfrac{\dfrac{y}{x}-1}{\dfrac{y}{x}+1}\\ y'=f(\dfrac{y}{x})$$

  Therefore, the ordinary differential equation is homogeneous.

  As $z=\dfrac{y}{x}$ , $f(z)=\dfrac{z-1}{z+1}$, we'll substitute into eq. 3 since it was proved, we optimize time.

• Orthogonal slope to circle.

  https://www.geogebra.org/calculator/qnzqyahg

• La intersección con el eje de las abscisas de la recta tangente a una curva en un punto cualquiera es siempre igual a la ordenada de dicho punto. Si una miembro de la familia pasa por el punto , determinar la ecuación de dicha curva.

  https://www.geogebra.org/calculator/mpbhtugf

Second order linear equation

Let $y_2$ and $y_1$ solutions then the general solution is

Uniqueness Theorem.

$\text{If } y''+Py'+Qy=F(x),y(a)=b,y'(a)=c\\ P,Q,F$ continuous on $I$ then $y(x)$ is unique.

Lineal dependency

Worked examples

• Let $ty''-\dfrac{6}{t}y=0$, assert $y_1(t)=t^3$ it is a solution, and then search $u(t)$ such that $y_2(t)=u(t)y_1(t)$.

  $$t(6t)-\dfrac{6}{t} (t^3)=0 \implies y_1=t^3 \text{ It is a solution.}\\ \text{We have }\\ y_c=c_1y_1+c_2y_2\\ y_2'=u'y_1+uy_1'\\ y_2''=u''y_1+u'y_1'+u'y_1'+uy_1''\\ \text{Substitute in ODE with } c_1=0,c_2=1\\ (u''y_1+2u'y_1'+uy_1'')-\dfrac{6}{t^2}(uy_1)=0\\ (t^3)u''+2u'(3t^2)+(u6t)-\dfrac{6}{t^2}(ut^3)=0\\ (t^3)u''+(6t^2)u'=0\\ u''+\dfrac{6}{t}u'=0\\ \text{Let }v=u'\implies v'=u''\\ \int \dfrac{v'}{v} dt=\int -\dfrac{6}{t}dt\\ v(t)=\dfrac{c}{t^6}\\ \dfrac{du}{dt}=\dfrac{c}{t^6}\\ u(t)=-\dfrac{c_1}{t^5}+c_2\\ \text{Thus }\\ y_2=(\dfrac{-c_1}{t^5}+c_2)(t^3)\\ y_2=\dfrac{-c_1}{t^2}+c_2t_3\\ y_c=c_1t^3+c_2\dfrac{1}{t^2}\\ Prove:(\dfrac{6t^5+6}{t^4})-\dfrac{6}{t^2}(t^3+\dfrac{1}{t^2})=0\\ W(y_1,y_2)=-5$$

• $\text{Euler-Cauchy equation}:t^2y''(t)-6y(t)=t^4$. Use $z=ln(t),t=e^z$

  $$\text{Let }y(t)=u(z(t))\text{ an ODE solution}\\ \text{Chain rule:}y'(t)=u'(z(t))z'(t)\\ y''(t)=u''(z(t))z'(t)z'(t)+u'(z(t))z''(t)\\ y''(t)=u''(z(t))\dfrac{1}{t^2}+u'(z(t))(\dfrac{-1}{t^2})=\dfrac{1}{t^2}(u''(z(t))-u'(z(t)))\\ \text{Substitute } t=e^z\\ u''(z(t))-u'(z(t))-6u(z(t))=e^{4z(t)}\\ u''(z)-u'(z)-6u(z)=e^{4z}\\ \text{Ansatz: }u_c(z)=e^{rz}\\ r^2-r-6=0\\ u_c(z)=c_1e^{3z}+c_2e^{-2z}\\ \text{Ansatz: }u_p(z)=Ae^{4z}\\ 16Ae^{4z}-4Ae^{4z}-6Ae^{4z}=e^{4z}\\ 16A-4A-6A=1\\ A=\dfrac{1}{6}\\ u_p(z)=\dfrac{1}{6}e^{4z} \\ \text{Thus, } y(t)=u(z)=u_c+u_p=c_1e^{3z}+c_2e^{-2z}+\dfrac{1}{6}e^{4z}\\ \text{Substitute }z=ln(t)\\ y(t)=c_1x^3+c_2\dfrac{1}{x^2}+\dfrac{1}{6}x^4$$

• Solution.

  $$mx''+cx'+kx=0\\ Ansatz:x(t)=e^{rt}\implies mr^2+cr+k=0\\ r=\dfrac{-c\pm \sqrt{p}}{2m}\\ p=c^2-4mk\\ Substitute:p=9^2-4*10*2=1\\ \text{As p>0, then }\\ a_0=\dfrac{-9+ \sqrt{1}}{2*10}=-0.4\\ a_1=\dfrac{-9- \sqrt{1}}{2*10}=-0.5\\ \text{Solution: }x(t)=c_1e^{-0.4t}+c_2e^{-0.5t}\\ \text{But }x(0)=c_1+c_2=0\\ x'(0)=-0.4c_1+-0.5c_2=5 \\ \text{By Gauss Jordan }\\ c_1=50,c_2=-50\\ \text{Particular solution: }x(t)=50e^{-0.4t}-50e^{-0.5t}$$

  $$lim_{t\rarr \infty} x(t)=lim_{t\rarr \infty} 50e^{-0.4t}-lim_{t\rarr \infty} 50e^{-0.5t}=0-0=0$$