5.6. Differential equations

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1 5.6. Differential equations The relationship between cause and effect in phsical phenomena can often be formulated using differential equations which describe how a phsical measure () and its derivative () depend on each other. Eamples are distance s(t) and velocit v(t) s (t) velocit v(t) and acceleration a(t) v (t) electrical charge Q(t) and amperage I(t) Q (t) work W(t) und power P(t) W (t) When () and () appear together in a single equation it is generall not possible to solve the equation to (). However most phsical applications lead to certain simple tpes of differential equations which can be solved with the special methods eplained below. In the general case onl approimate solutions are possible and two common algorithms for these will be presented too Separation of variables Separation of variables A differential equation of the form f() g() with initial value ( ) is called variables separable and can often be solved in the following wa:. Separation of variables: Get all epressions in on one side and all epressions in on the other side: g() f().. Integrate starting from the given initial value ( ) as lower bound: 3. Solve to. Caution: Steps and 3 ma not alwas work! Eample: g() Find the solution () to the differential equation () () 5 with initial value () 3,5 () () 5 use differential notation 5 separation of variables 5 integrate on both sides: 5 ln( 5),5 Solve integrals plug in [ln( 5) ln( 3 5)] simplif: ln() ln 5 put to the power of e 5 e () e 5 Eercises on differential equations Nos and simplif f ().

2 5.6.. Homogenous differential equations A differential equation of the form d(v ) using the substitution v with differential equation with respect to v then becomes variables. Eample: Find the general solution () to the differential equation () The differential equation is homogenous with right hand side The usual substitution v v v with f f (v) v is called homogenous and can be converted into a variables separable form + v + v f(v) f (v) f + v and f(v) v v v v with, >. + v according to the product rule. The v. ields the variables separable form separation of variables integration with indefinite lower limits > ln v ln() + c put to the power of e v e c solve to v v resubstitute v e c and can be solved b separating the and multipl with c e Eercises on differential equations Nos Linear differential equations A differential equation of the form integrating factor I() e I() P() + I() P() () I() Q() I() + di d(i ) + P() () Q() is called linear and can be solved b multipling with the. Since this integrating factor satisfies di di plug in I() P() we obtain I() P() () I() Q() appl product rule I() P() integrate I() () I() P() solve to () () I() I() P().

3 Eample: Find the solution () to the differential equation cos() + sin() (cos()) with (). Dividing b cos() ields a linear differential equation () + tan() () cos() with integrating factor I() P() e tan() e cos(). Note that the frequentl occurring integral tan() ( sin()) cos() ln ln cos() ln cos() is not included in the formular booklet (!) Multipling with I() obtains cos() () + sin() () product rule (cos()) d () cos() () cos() integration + c solve to () cos() ( + c) plug in initial value to obtain c cos() ( + c) solve to c c cos() ( + ) Eercises on differential equations Nos 7 and 8 state special solution Slope fields and the Euler method Slope fields The differential equation f(, ) gives the gradient or slope of the function () in an given point ( ). The slope field is constructed b small arrows with gradient f(,) on each point ( ). To save time and effort lines of constant gradient can be used to replace the arrows. The equations such an isocline with gradient c has the equation c f(,). Eample a) Construct the slope field of + for all points with integer coordinates in the domain 3, 3 b) Draw the isoclines of + in the domain 3, 3 for slopes c { 3,,, 3} 3

4 Solution We use a table to plug in all 49 combinations of integer coordinates in the given domain. For eample the point ( 3) has the slope c 3 c c c c c c 3 The isoclines c + ± c can be obtained b translation of in direction of the -ais c steps to the right The Euler method To find an approimate solution for the differential equation f(, ) with given initial value ( ) and step length Δ proceed as follows: Start from the given point ( ), calculate the gradient m f(, ) and add the corresponding values Δ and Δ m Δ to obtain the net point ( ) ( + Δ + Δ). Using the new point we repeat the above described steps to get a point sequence ( i i ) i which approimates the ideal solution (). Eample: Using Euler s method with step length of,5, approimate the solution,, of the differential equation + when, given that when. Solution It is convenient to organize the results in a table: m,,,5,5,6875,5,4,4783,75,54,374,,6349 Result (),63 Eercises on differential equations Nos9 and 4

5 Using Talor polnomials Basicall the differential equation f(, ) gives a formula for the derivative of the function. Hence it can be used ''( ) repeatedl to develop the Talor series () ( ) + ( ) ( ) + ( ) +. The coefficients are obtained as! follows: ( ) ; ( ) f(, ), ( ) f (, ), Eample Find a Talor polnomial of degree 3 to approimate the solution of the differential equation + with () and use this polnomial to find an approimation of (,) to 4 decimal points. Solution B implicit differentiation we obtain () ( + ) + and () ( + ) + +. We plug in () ; () +, () + and () The Talor polnomial is () () + () ( ) + ''()! The approimate value at, is (,), +, +,,7. 3 ( ) + '''() ( ) 3 ( ) + ( ) + 3! 3 ( )3. Eercises on differential equations No 5