# application of second order differential equation in chemistry

If y is a quantity depending on x, a model may be based on the following assumptions: The differential decrease of the variable y is proportional to a differential increase of the other variable, here x, i.e. As we discussed above, we’ll assume the solution is $$y(x)=e^{\alpha x}$$, and we’ll determine which values of $$\alpha$$ satisfy this particular differential equation. Whereas the response curve shown in B can be modelled with a first-order differential equation; the curve shown in C needs higher-order differential equations [5]. κ is a value specific for the particles and specific for the photon energy E 9. Let us suppose that the double layer is initially completely discharged (the electrode rests at the potential of zero charge), i.e. - 165.227.152.114. However, not each hitting leads to an absorption event (capture of a photon). We found two independent solutions to the differential equation, and now we will claim that any linear combination of these two independent solutions ($$c_1 y_1(x)+c_2 y_2(x)$$) is also a solution. Let’s see how it works with an example. First-order differential equations in chemistry, $$\frac{{{\text{d}}y}}{{{\text{d}}x}} + ay = 0$$, $$\frac{{{\text{d}}y}}{y} = - a{\text{d}}x$$, $$\int\limits_{{y_{0} }}^{y} {\frac{{{\text{d}}y}}{y}} = - a\int\limits_{0}^{x} {{\text{d}}x}$$, $$\lambda = {{hc_{\text{light}} } \mathord{\left/ {\vphantom {{hc_{\text{light}} } {E_{\text{photon}} }}} \right. first-order differential equation. The second initial condition is \(y'(0)=-1$$. Thus follows the ordinary linear homogeneous first-order differential equation: The characteristics of an ordinary linear homogeneous first-order differential equation are: (i) there is only one independent variable, i.e. 21 using the definition $$\frac{{a_{1} }}{{a_{2} }} = \tau$$ (Greek letter tau), i.e. For example: Find the solution of $$y''(x) -5y'(x) +4 y(x) = 0$$ subject to initial conditions $$y(0)=1$$ and $$y'(0)=-1$$. The response is then described by. 4 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS FORCED VIBRATIONS Suppose that, in addition to the restoring force and the damping force, the motion of the spring is affected by an external force . 54). (This is equal to solving the differential equation $$\frac{{{\text{d}}q'}}{{{\text{d}}t}} + \frac{1}{RC}q' = 0$$). For a comprehensive overview, see [9, 10]. \kern-0pt} {T_{1} }} \), $${1 \mathord{\left/ {\vphantom {1 {T_{2} }}} \right. In electrochemistry, such electrode is called an ideally polarizable electrode. $y'(x)=4c_1e^{4x}+c_2e^x\rightarrow y'(0)=4c_1+c_2=-1 \nonumber$. 2: Integration of Eq. Facts On File Inc, New York, p 91/92, Tebbutt P (1998) Basic mathematics for chemists, 2nd edn. The response curve shown in Fig. $y(x)=-\frac{2}{3}e^{4x}+\frac{5}{3}e^x \nonumber$. In general, all chemical reactions can be described mathematically by first-order differential equations. Following the same methodology we discussed for the previous example, we assume \(y(x)=e^{\alpha x}$$, and use this expression in the differential equation to obtain a quadratic equation in $$\alpha$$: $\alpha_{1,2}=\frac{3\pm \sqrt{(-3)^{2}-4\times 9/2}}{2}=\frac{3\pm \sqrt{-9}}{2} \nonumber$. http://tinyurl.com/mpl69ju. The origin of time constants is a very complex topic needing extensive explanations. $$t_{90\% }$$ and $$t_{99\% }$$, as the signal reached after that time is often regarded as a good estimate of the true signal value. An examination of the forces on a spring-mass system results in a differential equation of the form $mx″+bx′+kx=f(t), \nonumber$ where mm represents the mass, bb is the coefficient of the damping force, $$k$$ is the spring constant, and $$f(t)$$ represents any net external forces on the system. The different sensor behaviours shown in B and C can be modelled with the help of different differential equations. 5b can be modelled as follows: $$z$$ is a time-dependent quantity for which we write the first-order differential equation. Chemical reaction kinetics is the study of rates of chemical processes (reactions). \kern-0pt} {RC}}}} ) $$,$$ I = \frac{{{\text{d}}q}}{{{\text{d}}t}} = - \frac{1}{RC}( - C \cdot \Delta E) \cdot {\text{e}}^{{ - {t \mathord{\left/ {\vphantom {t {RC}}} \right. the rate of decay. The double layer has the property of a capacitor, as charge can be stored on both sides, always equal amounts with opposite signs. Felix Bloch [11] described two different relaxation processes with (i) the spin–lattice or longitudinal relaxation time $$T_{1}$$, describing the relaxation in the direction of the external magnetic field B 13 has a restricted range of validity: it is a good description of real systems only at low concentrations. This means that $$e^{-k_1 x/2}$$ is a solution, but this creates a problem because we need another independent solution to create the general solution from a linear combination, and we have only one. Time response of a sensor when it can be modelled with a first-order differential equation.

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