Differential equations arise in many problems in physics, engineering, and other sciences. The following examples show how to solve differential equations in a few simple cases when an exact solution exists.
- Example Of Equation
- Linear Differential Equation Example Pdf
- Bernoulli Differential Equation Examples Pdf
- 4Second-order linear ordinary differential equations
Separable first-order ordinary differential equations[edit]
Equations in the form dydx=f(x)g(y){displaystyle {frac {dy}{dx}}=f(x)g(y)} are called separable and solved by dyg(y)=f(x)dx{displaystyle {frac {dy}{g(y)}}=f(x)dx} and thus ∫dyg(y)=∫f(x)dx{displaystyle int {frac {dy}{g(y)}}=int f(x)dx}. Prior to dividing by g(y){displaystyle g(y)}, one needs to check if there are stationary (also called equilibrium)solutions y=const{displaystyle y=const} satisfying g(y)=0{displaystyle g(y)=0}.
Separable (homogeneous) first-order linear ordinary differential equations[edit]
MATLAB Tutorial on ordinary differential equation solver (Example 12–1) Solve the following differential equation for co-current heat exchange case and plot X, Xe, T, Ta, and -rA down the length of the reactor (Refer LEP 12–1, Elements of chemical reaction engineering, 5th edition) Differential equations. figure out this adaptation using the differential equation from the first example. Then, if we are successful, we can discuss its use more generally.! Example 4.3: Consider the differential equation dy dx − x2y2 = x2. In example 4.1 we saw that this is a separable equation, and can be written as dy dx = x2 1 + y2. It is the same concept when solving differential equations — find general solution first, then substitute given numbers to find particular solutions. Let’s see some examples of first order, first degree DEs. Find the general solution for the differential equation `dy + 7x dx = 0` b. Find the particular solution given that `y(0)=3`.
A separable linearordinary differential equation of the first order must be homogeneous and has the general form
dydt+f(t)y=0{displaystyle {frac {dy}{dt}}+f(t)y=0}
where f(t){displaystyle f(t)} is some known function. We may solve this by separation of variables (moving the y terms to one side and the t terms to the other side),
dyy=−f(t)dt{displaystyle {frac {dy}{y}}=-f(t),dt}
Since the separation of variables in this case involves dividing by y, we must check if the constant function y=0 is a solution of the original equation. Trivially, if y=0 then y’=0, so y=0 is actually a solution of the original equation. We note that y=0 is not allowed in the transformed equation.
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We solve the transformed equation with the variables already separated by Integrating,
ln|y|=(−∫f(t)dt)+C{displaystyle ln |y|=left(-int f(t),dtright)+C}
where C is an arbitrary constant. Then, by exponentiation, we obtain
y=±e(−∫f(t)dt)+C=±eCe−∫f(t)dt{displaystyle y=pm e^{left(-int f(t),dtright)+C}=pm e^{C}e^{-int f(t),dt}}.
Here, 0}”>eC>0{displaystyle e^{C}>0}, so ±eC≠0{displaystyle pm e^{C}neq 0}. But we have independently checked that y=0 is also a solution of the original equation, thus
y=Ae−∫f(t)dt{displaystyle y=Ae^{-int f(t),dt}}.
with an arbitrary constant A, which covers all the cases. It is easy to confirm that this is a solution by plugging it into the original differential equation:
dydt+f(t)y=−f(t)⋅Ae−∫f(t)dt+f(t)⋅Ae−∫f(t)dt=0{displaystyle {frac {dy}{dt}}+f(t)y=-f(t)cdot Ae^{-int f(t),dt}+f(t)cdot Ae^{-int f(t),dt}=0}
Some elaboration is needed because ƒ(t) might not even be integrable. One must also assume something about the domains of the functions involved before the equation is fully defined. The solution above assumes the real case.
If f(t)=α{displaystyle f(t)=alpha } is a constant, the solution is particularly simple, y=Ae−αt{displaystyle y=Ae^{-alpha t}} and describes, e.g., if 0}”>α>0{displaystyle alpha >0}, the exponential decay of radioactive material at the macroscopic level. If the value of α{displaystyle alpha } is not known a priori, it can be determined from two measurements of the solution. For example,
dydt+αy=0,y(1)=2,y(2)=1{displaystyle {frac {dy}{dt}}+alpha y=0,y(1)=2,y(2)=1}
gives α=ln(2){displaystyle alpha =ln(2)} and y=4e−ln(2)t=22−t{displaystyle y=4e^{-ln(2)t}=2^{2-t}}.
Non-separable (non-homogeneous) first-order linear ordinary differential equations[edit]
First-order linear non-homogeneous ODEs (ordinary differential equations) are not separable. They can be solved by the following approach, known as an integrating factor method. Consider first-order linear ODEs of the general form:
dydx+p(x)y=q(x){displaystyle {frac {dy}{dx}}+p(x)y=q(x)}
The method for solving this equation relies on a special integrating factor, μ:
μ=e∫x0xp(t)dt{displaystyle mu =e^{int _{x_{0}}^{x}p(t),dt}}
We choose this integrating factor because it has the special property that its derivative is itself times the function we are integrating, that is:
Example Of Equation
dμdx=e∫x0xp(t)dt⋅p(x)=μp(x){displaystyle {frac {d{mu }}{dx}}=e^{int _{x_{0}}^{x}p(t),dt}cdot p(x)=mu p(x)}
Multiply both sides of the original differential equation by μ to get:
μdydx+μp(x)y=μq(x){displaystyle mu {frac {dy}{dx}}+mu {p(x)y}=mu {q(x)}}
Because of the special μ we picked, we may substitute dμ/dx for μp(x), simplifying the equation to:
μdydx+ydμdx=μq(x){displaystyle mu {frac {dy}{dx}}+y{frac {d{mu }}{dx}}=mu {q(x)}}
Using the product rule in reverse, we get:
ddx(μy)=μq(x){displaystyle {frac {d}{dx}}{(mu {y})}=mu {q(x)}}
Integrating both sides:
μy=(∫μq(x)dx)+C{displaystyle mu {y}=left(int mu q(x),dxright)+C}
Finally, to solve for y we divide both sides by μ{displaystyle mu }:
y=(∫μq(x)dx)+Cμ{displaystyle y={frac {left(int mu q(x),dxright)+C}{mu }}}
Since μ is a function of x, we cannot simplify any further directly.
Second-order linear ordinary differential equations[edit]
A simple example[edit]
Suppose a mass is attached to a spring which exerts an attractive force on the mass proportional to the extension/compression of the spring. For now, we may ignore any other forces (gravity, friction, etc.). We shall write the extension of the spring at a time t as x(t). Now, using Newton’s second law we can write (using convenient units):
md2xdt2+kx=0,{displaystyle m{frac {d^{2}x}{dt^{2}}}+kx=0,}
where m is the mass and k is the spring constant that represents a measure of spring stiffness. For simplicity’s sake, let us take m=k as an example.
If we look for solutions that have the form Ceλt{displaystyle Ce^{lambda t}}, where C is a constant, we discover the relationship λ2+1=0{displaystyle lambda ^{2}+1=0}, and thus λ{displaystyle lambda } must be one of the complex numbersi{displaystyle i} or −i{displaystyle -i}. Thus, using Euler’s formula we can say that the solution must be of the form:
x(t)=Acost+Bsint{displaystyle x(t)=Acos t+Bsin t}
See a solution by WolframAlpha.
Linear Differential Equation Example Pdf
To determine the unknown constants A and B, we need initial conditions, i.e. equalities that specify the state of the system at a given time (usually t = 0).
For example, if we suppose at t = 0 the extension is a unit distance (x = 1), and the particle is not moving (dx/dt = 0). We have
x(0)=Acos0+Bsin0=A=1,{displaystyle x(0)=Acos 0+Bsin 0=A=1,}
and so A = 1.
x′(0)=−Asin0+Bcos0=B=0,{displaystyle x’(0)=-Asin 0+Bcos 0=B=0,}
and so B = 0.
Therefore x(t) = cos t. This is an example of simple harmonic motion.
See a solution by Wolfram Alpha.
A more complicated model[edit]
The above model of an oscillating mass on a spring is plausible but not very realistic: in practice, friction will tend to decelerate the mass and have magnitude proportional to its velocity (i.e. dx/dt). Our new differential equation, expressing the balancing of the acceleration and the forces, is
Bernoulli Differential Equation Examples Pdf
md2xdt2+cdxdt+kx=0,{displaystyle m{frac {d^{2}x}{dt^{2}}}+c{frac {dx}{dt}}+kx=0,}
where c{displaystyle c} is the damping coefficient representing friction. Again looking for solutions of the form Ceλt{displaystyle Ce^{lambda t}}, we find that
mλ2+cλ+k=0.{displaystyle mlambda ^{2}+clambda +k=0.}
This is a quadratic equation which we can solve. If c2<4km{displaystyle c^{2}<4km} there are two complex conjugate roots a ± ib, and the solution (with the above boundary conditions) will look like this:
x(t)=eat(cosbt−absinbt){displaystyle x(t)=e^{at}left(cos bt-{frac {a}{b}}sin btright)}
Let us for simplicity take m=1{displaystyle m=1}, then 0<c=−2a{displaystyle 0<c=-2a} and k=a2+b2{displaystyle k=a^{2}+b^{2}}.
The equation can be also solved in MATLAB symbolic toolbox as
although the solution looks rather ugly,
This is a model of a damped oscillator. The plot of displacement against time would look like this:
which resembles how one would expect a vibrating spring to behave as friction removes energy from the system.
Linear systems of ODEs[edit]
The following example of a first order linear systems of ODEs
y1′=y1+2y2+t{displaystyle y_{1}’=y_{1}+2y_{2}+t}y2′=2y1−2y2+sin(t){displaystyle y_{2}’=2y_{1}-2y_{2}+sin(t)}
can be easily solved symbolically using numerical analysis software.
See also[edit]
Bibliography[edit]
- A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for Ordinary Differential Equations, 2nd Edition, Chapman & Hall/CRC Press, Boca Raton, 2003; ISBN1–58488–297–2.
External links[edit]
- Ordinary Differential Equations at EqWorld: The World of Mathematical Equations.
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