> <\body> \ |>>> General form\ <\equation*> F|)>=0 <\example> The motion of the object is governed by the Newton's law. Let > be the velocity of the object at time . <\equation*> m*v=m*g-\*v, where is the mass of the object, is the gravitational constant, and > is the coefficient of air resistant force. This is a first order linear ODE. <\example> Find the general solutions of <\equation*> u+u=2 idea: combine +u> into the derivative of another function.\ Consider the product rule of differentiation: <\equation*> =f*g+f*g. Let ,g=e>, then <\equation*> *e|]>=u*e+u*e=e*+u|)>. Now, we multiply the original equation by >: <\equation*> +u|)>*e=2*e\*e|]>=2*e\u*e=2*e*\t=2*e+c Divide by >: <\equation*> u=e*+c|]>=2+c*e. Consider the 1st order linear ODE (standard form): <\equation*> +p*u=q.> Multiply left side by >: <\equation*> +p*u|]>*\=u*\+p*u*\. We want <\equation*> \=g,p*\=g i.e. <\equation*> \=p**\\=ep*\t>>. Check (by the chain rule and fundamental theorem of calculus) <\equation*> \=ep*\t>*p=\*p. Derivation directly: <\equation*> \=p**\\|\>=p\|]>=p <\equation*> \ln*\=p*\t\\=ep*\t.> So the original ODE becomes <\equation*> *u|]>=\*q\\*u=\*q*\t+c <\equation*> \=>*\*q*\t+c|]>>, where <\equation*> =ep*\t>> is called the . <\example> Find the of <\equation*> |)>*+2*t*y=4*t Rewrite it in the standard form <\equation*> y+>*y=>. Here =>>. Then find the integrating factor >: <\equation*> \=ep*\t>=e|)>>=4+t. So the solution is <\equation*> y=>*|)>*>*\t+c|)>=>*+c|)> <\example> Solve the +2*y=4*t>>|y=2>>>>> First rewrite the equation into the standard form: <\equation*> y+*y=4*t. Find the integrating factor: <\equation*> \=e*\t>=e>=e>=t. The general solution is <\equation*> y=>* t*4*t*\t+c|]>=>*+c|]>=t+>. Plugging the initial condition: <\equation*> y=2\1+=2\. The solution of the initial value problem is <\equation*> y=t+>. \; <\example> Solve <\equation*> =|1-y>. This equation is nonlinear. We can separate the variables and as follows <\equation*> |)>*\y=x*\x\|)>*\y=x*\x <\equation*> \y-|3>=|3>+c This is an example of . <\definition> An ODE in the form of <\equation*> y|\x>=|g> is called We can solve it as follows <\equation*> f*\x=g*\y <\example> Solve the initial value problem <\equation*> =+4*x+2|2*>,y=-1 \ <\equation*> +4*x+2|)>*\x=2**\y <\equation*> \x+2*x+2*x=y-2*y+c. Plugging the initial condition <\equation*> y=-1\0=1+2+c\c=-3. The solution to the initial value problem is <\equation*> x+2*x+2*x=y-2*y-3. <\question> What is the domain and range of the solution? <\example> Recall the differential equation for continuous compound interests: <\equation*> u=r*u. Note this equation is both linear and separable. As a separable equation, we have <\equation*> u|\t>=r*u\u|u>=r*\t\u|u>=r*\t\ln =r*t+c <\equation*> \=c*e\u=c*e. If the initial condition is =u>,then we find >. <\exercise> Solve the equation by the method of integrating factor. Suppose =c> is an solution of some ODE. Taking x> on both sides of the solution. <\equation*> x>*\=x>*c\\|\x>>+\|\y>>*y|\x>=0\>+>*y=0, where <\equation*> M=\\,N=\*\. \; <\example> Solve >+*y=0>. <\answer*> Guess the solution. Let =x+y*x>. Then <\equation*> \=2*x+y,\=2*x*y. So <\equation*> 0=\+\*y=x>*\ So the solution is <\equation*> \=c. <\definition*> An ODE of the form <\equation*> M+N*y=0M*\x+N*\y=0 is called if there exists > such that <\equation*> \=M,\=N. The solution of the equation is <\equation*> \=c, where is an arbitrary constant. <\theorem> Suppose an ODE can be written in the form <\equation> M+N*y=0M*\x+N*\y=0 where the functions > and > are all continuous in the rectangular region \>. Then Eq. (1) is an exact differential equation <\equation*> M=N,\\R. <\proof> \>. Suppose Eq. (1) is exact. Then there exists a > such that <\equation*> \=M,\=N. Then <\equation*> M=\,N=\. Since ,N> are continuous, we have > and > are continuous. So\ <\equation*> \=\. i.e. <\equation*> M=N. \> Suppose =N>. We want to find a function > such that =M> and =N>. Let <\equation*> \=M*\x+h. Then =M>, and <\equation*> \=\M*\x+h. We want =N>, that is <\equation*> h=N-\M*\x. We need the RHS to be independent of . That is <\equation*> |\x> -\ *M*\x|]>=0. Let's check: <\eqnarray*> |\x>-\*M*\x|]>>||-\ \M*\x=N-M=0.>>>> \; \; <\example> Solve the ODE <\equation*> |)>+*e-1|)>*y=0. \ <\eqnarray*> >||>>|>||>>>> So =N>, and the equation is exact. Next, let <\equation*> \=M d*x=y*+2*x*e d*x=y*+x*e+h. Then <\equation*> \=sin x+x*e+h=N=sin x+x*e-1 <\equation*> \h=-1\h=-y. So the solution is <\equation*> \=y*+x*e-y=c. <\exercise*> Solve the above equation, but using =N d*y+h> first. <\question*> What is the relationship between separable and exact equations? Sometimes we can multiply a function to a non-exact equation to make it exact. Take a function \0>,\ <\eqnarray*> d*x+N d*y>||>| d*x+N d*y|]>>||>|*M d*x+\*N d*y>||>| d*x+ d*y>||>>> where =\*M,=\*N>. Then let <\equation*> =\*M+\*M,=\*N+\*N. We want =>, i.e. <\equation*> \*M+\*M=\*N+\*N. Let's choose > such that =0>. Then the above equation reduces to <\equation*> \*M=\*N+\*N\\=-N|N>*\. If the function -N|)>/N> is a function of only, then we can solve > as a separable equation. Here > is called an integrating factor. <\wide-block> <\example> Solve the ODE <\equation*> |)>++x*y|)>*y=0. It's first order, nonlinear, and not separable. Check if it's exact: <\equation*> M=3*x+2*y,N=2*x+y. Not exact!. Next, try integrating factors. <\equation*> -N|N>=+x*y>= is a function of only! Let <\equation*> \=*\\\=x. Then multiply > to the original equation: <\equation*> x*|)>+x*+x*y|)>*y=0 <\equation*> *y+x*y|)>++x*y|)>*y=0 Double check the new equation is exact! Then solve it as usual (exercise). >>> Similarly, if -M|)>/M> is a function of only, then we can use the integrating factor > solving <\equation*> \=-M|M>*\. Consider the first order ODE: <\equation*> y=f Draw small arrows as a vector |)>> at many points > Online plotter: <\example*> Consider <\equation*> y=> \; <\theorem*> Consider the initial value problem <\equation*> y+p*y=q,y|)>=y. If are continuous on an interval > containing >, then the IVP has a unique solution on . <\example*> Consider <\equation*> t*y+2*y=4*t,y=2. Solve it by integrating factors,\ <\equation*> y+*y=4*t\\=exp d*t|]>=t. <\equation*> y=>4*t d*t+c|]>=>+c|]>=t+>. Plugging =2>, we obtain . The solution is <\equation*> y=t+>. Now, =,q=4*t>. So are continuous in ,0|)>\|)>>. But |)>> only, so we know from the theorem the IVP has a unique solution in |)>>, which is <\equation*> y=t+>,t\|)>. If the initial condition is changed to =2>, then the solution is <\equation*> y=t+>,t\,0|)>. <\theorem*> Consider the initial value problem <\equation*> y=f,y|)>=y. If and f> are continuous on a rectangular domain \> containing the point ,y|)>>. Then the IVP has a unique solution in some interval containing >. <\example*> Consider the IVP. <\equation*> =+4*x+2|2*>,y=-1 It is separable. Let's solve it first,\ <\equation*> 2*d*y=+4*x+2|)>*d*x\y-2*y=x+2*x+2*x+c <\equation*> y=-1\c=3. The solution is <\equation*> y-2*y=x+2*x+2*x+3 <\equation*> y=+2*x+2*x+3|)>>|2>=1\+2*x+2*x+4>=1-+2*x+2*x+4>. Here and f> are continuous everywhere except . <\example*> Consider <\equation*> y=y,y=0*0|)> First, let's solve it as a separable equation. <\equation*> y d*y=d*t\*y=t+c Plugging =0> yields . So <\equation*> y=\*t|)> are two solutions. In addition <\equation*> y=0 is also a solution. In fact, we have infinitely many solutions defined as <\equation*> y=|t>>||*>*t-t|)>|]>,>|t>>>>>,y=|t>>|*|)>|]>,>|t>>>>> for any \0>. ( check is continuous and differentiable at >.) |gr-frame|>|gr-geometry||gr-color|red|gr-auto-crop|true|||>>|||>>||>>|||>>||>>|||>>||>>|||>>||>>||>>|||>>|>>> In fact,\ <\equation*> f=y,\ f=*y. So f> is discontinuous near >. So there exists no rectangle containing > such that f> are both continuous in . So we can't gaurantee the existence and uniqueness of solution for the IVP. <\note> One may not be able to find all solutions to nonlinear equations using one method.\ <\example*> Consider <\equation*> =y,y=y\0 We can first solve it as a separable eqn. <\equation*> y*d*y=d*t\-y=t+c\y=-\y=->>. Now ,\ f=2*y> are continuous everywhere. However, the solution is not defined for every . For example, if \0>,then the solution is defined only in ,>|)>>. <\equation*> m*v=m*g-\*v, where is the velocity, > are constants.\ <\itemize> Analyze the solutions using direction field. Solve it by integrating factors. <\equation*> v+|m>*v=g integrating factor <\equation*> \=e\/m>=e|m>*t> <\equation*> v=e|m>*t>*g*e|m> t>*d*t+c|]>=e|m>*t>*>*e|m>*t>+c|]>=>+c*e|m>*t>. If the initial condition is =v>. Then -g*m/\>. So the solution of the IVP is <\equation*> v=>+->|]>*e|m>*t>. So <\equation*> lim\> v=>. All other solutions converge to the > as \>. This equilibrium solution is a one. Assume the annual interest rate is . The continuous rate of deposit/withdrawal is . Then the ODE model for the total balance > is <\equation*> u=r*u+k. integrating factor <\equation*> \=e <\equation*> u=e*k*e+c|]>=e**e+c|]>=-+c*e. If the initial condition is =u>, then +k/r>. So the solution of the IVP is <\equation*> u=-++|)>*e. The equilibrium solution is >, and it is an one since all other solutions diverge from it as \>. <\equation*> y=r*y, The solution is <\equation*> y=y*e where =y>.\ <\itemize> If 0>, we have exponential growth If 0>, we have exponential decay, such as radioactive decay. <\equation*> y=*y. Note that the right-hand-side depends on only. In general, ODE of the form <\equation*> y=f is called . There are two equilibrium solutions <\equation*> y=0,y=. From the direction field we can tell the equilibrium solution is unstable, while the solution > is stable. <\wide-centered> |png>|0.4par|||> Now let's solve the equation: <\equation*> *y>=d*t\*y>=d*t\ <\equation*> *y>=+=+B*y|y*>\1=A*+B*y <\equation*> y=0\A=1/r,y=r/a\B=a/r So <\equation*> *y>=+*d*y=*ln +|)>*ln =*ln -*ln <\equation*> =*ln |>=d*t=t+c <\equation*> \|>=e>=c*e\=c*e. <\equation*> \y=|1+a*c*e>=+a*c>=*e+a>. Suppose the initial condition is =y>, then\ <\equation*> c=|r-a*y>\y=|y>*e+a>=|a*y+|)>*e>=|y+|)>*e>>>>>>, where >. Note that =*y=r*|)>*y> <\enumerate> If y\K>, then > is an increasing function, and \> y=K>, but \K> for all 0>. Moreover, \> y=0>, and <\equation*> y= = f=f*=f*f, where <\equation*> f=r*|)>*y,f=r*|)> <\enumerate> If y\>, then \0>, so the graph is concave up. If \y\K>, then \0>, so the graph is concave down. If \K>, then > is an decreasing function, and \> y=K>, but \K> for all 0>. Moreover, \> y=0>, and \0> for all . Consider a general 1st order ODE <\equation*> y=f. Take ,y|)>>, then <\equation*> y|)>=f,y|)> <\equation*> y|)>=lim0> +h|)>-y|)>|h>\|)>-y|)>|t-t> if -t|\|>> is small. So <\equation*> y|)>\y|)>+-t|)>*y|)>=y|)>+-t|)>*f,y|)> Let <\equation*> y=y+-t|)>*f,y|)> So \y|)>>. Repeat this process, we obtain an algorithm: For a sequence of ,t,t,\> <\equation*> =y+-t|)>*f,y|)>>>>>> This sequence of ,y,y,\> is an approximation of the true values |)>,y|)>,y|)>,\> \; \; \; > <\initial> <\collection> > <\references> <\collection> > > > > > > > > > > > > > > > > > > <\auxiliary> <\collection> <\associate|toc> |math-font-series||font-size||Chapter 2: First Order Differential Equations> |.>>>>|> |math-font-series||1Method of integrating factors> |.>>>>|> |math-font-series||2Separable equations> |.>>>>|> |math-font-series||3Exact Equations> |.>>>>|> |3.1Motivation and definition |.>>>>|> > |3.2Theorem and method |.>>>>|> > |3.3Integrating factors |.>>>>|> > |math-font-series||4Direction fields> |.>>>>|> |math-font-series||5The Existence and Uniqueness Theorem> |.>>>>|> |5.1Linear equations |.>>>>|> > |5.2Nonlinear equations |.>>>>|> > |math-font-series||6Applications> |.>>>>|> |6.1Falling object in the air |.>>>>|> > |6.2Compound interest with deposits/withdrawals |.>>>>|> > |6.3Population dynamics |.>>>>|> > |6.3.1Exponential growth |.>>>>|> > |6.3.2Logistic growth |.>>>>|> > |math-font-series||7Euler's method> |.>>>>|>