Elementary Differential Equations - …

1 Elementary Differential Equations First-order linear Differential Equations and solutions. Numerical solutions. Reference: E. Kreyszig,Advanced Engineering Mathematics, 5thed., Wiley, Linear Differential Equations General form takes:y +a(x)y=r(x),wherey =dy/dxandyis a function ofx. It is first order sincethe highest derivative is first order (y ). It is linear, because it is alinear function ofy andy(no terms likey2ory y, etc.). Homogeneous, whenr(x)=0. In this case, the solution can befound easily throughseparation of variables. Nonhomogeneous, whenr(x)6=0. The solution in this case isa bit more involved, and several different approaches 1st-order Linear Diff. Start withy +a(x)y= 02. Subtracta(x)yfrom both sides:y = a(x)y3. Divide both sides byy:1ydydx= a(x)4. Multiply both sides withdx:1ydy= a(x)dx5. Integrate:Z1ydy=Z a(x)dxlny= Za(x)dx+c6. Applyexp( ):y= exp Za(x)dx+c y= exp Za(x)dx exp(c)y=Cexp Za(x)dx This calledseparation of 1st-order Diff.

2 Start withy +a(x)y=r(x)2. Letyhbe the solution to the homogeneous case (wherer(x) = 0), so thaty h+a(x)yh= 03. For now, assumey(x) =yh(x)u(x)4. Plug in 3 to 1:d(yh(x)u(x))dx+a(x)yh(x)u(x)=r(x)y h(x)u(x)+yh(x)u (x)+a(x)yh(x)u(x)=r(x)u(x)(y h(x)+a(x)yh(x))|{z}This is 0, from step 2.+yh(x)u (x)=r(x)yh(x)u (x)=r(x)4 Nonhomogeneous 1st-order Diff. Continuing from:yh(x)u (x) =r(x)6. divide both sides byyh(x):dudx=r(x)yh(x)7. Multiply both sides withdx:du=r(x)yh(x)dx8. Integrate:Zdu=Zr(x)yh(x)dxu=Zr(x)yh(x)dx +c9. Plug this in toy(x) =yh(x)u(x)to get:y(x) =yh(x) Zr(x)yh(x)dx+c This calledvariation of Solution of Diff. Eq. (not so sophisticated)1. Givenf (x)and the initial conditionf(0)2. Take the Taylor series expansion off(x)arounda:f(x) =f(a) +f (a)(x a) +12!f (a)(x a)2+..3. Drop higher order terms:f(x) f(a) +f (a)(x a)4. Setx=t+ tanda=tfor a small t:f(t+ t) f(t) +f (t)(t+ t t) =f(t) +f (t) t5.

3 Starting withf(0), the approximate values f( )becomes: f( t) =f(0) +f (0) t f(2 t) = f( t) +f ( t) t f(3 t) = f(2 t) +f (2 t) tand so