ODE No. 88

  1. Problem in Latex
  2. Mathematica input
  3. Maple input

ce2ax4ay(x)b+2y(x)3y(x)2=0 Mathematica : cpu = 0.309628 (sec), leaf count = 2831

{{y(x)2(2a4a23b2a2a2+4a43a2ba2+13a4a23b2a24a24a43a2b4a2aa4a23b2a22a2+4a43a2b2a2+1ba4a23b2a24a24a43a2b4a21(a4a23b2a24a24a43a2b4a2)ca4a23b2a24a24a43a2b4a2e2axJ4a43a2b2a2(3bce2axb2a)Γ(4a43a2b2a2+1)(e2axb)a4a23b2a24a24a43a2b4a212a4a23b2a2a2+4a43a2ba223a4a23b2a24a24a43a2b4a2+12a4a43a2b2a2a4a23b2a22a2ba4a23b2a24a24a43a2b4a212ca4a23b2a24a24a43a2b4a2+12e2ax(J4a43a2b2a21(3bce2axb2a)J4a43a2b2a2+1(3bce2axb2a))Γ(4a43a2b2a2+1)(e2axb)a4a23b2a24a24a43a2b4a212+c1(22a24a23baa24a43a2ba2232a24a23ba4a2+4a43a2b4a2+12b2a24a23ba4a2+4a43a2b4a212c2a24a23ba4a2+4a43a2b4a2+12e2ax(e2axb)2a24a23ba4a2+4a43a2b4a212(J4a43a2b2a21(3bce2axb2a)J14a43a2b2a2(3bce2axb2a))Γ(14a43a2b2a2)a2a24a23ba2a24a43a2b2a222a24a23baa24a43a2ba2+132a24a23ba4a2+4a43a2b4a2b2a24a23ba4a2+4a43a2b4a21(2a24a23ba4a2+4a43a2b4a2)c2a24a23ba4a2+4a43a2b4a2e2ax(e2axb)2a24a23ba4a2+4a43a2b4a21J4a43a2b2a2(3bce2axb2a)Γ(14a43a2b2a2)a2a24a23ba2a24a43a2b2a2+1))3(22a24a23baa24a43a2ba232a24a23ba4a2+4a43a2b4a2b2a24a23ba4a2+4a43a2b4a2c2a24a23ba4a2+4a43a2b4a2(e2axb)2a24a23ba4a2+4a43a2b4a2J4a43a2b2a2(3bce2axb2a)c1Γ(14a43a2b2a2)a2a24a23ba2a24a43a2b2a2+24a43a2ba2a4a23b2a2a23a4a23b2a24a24a43a2b4a2ba4a23b2a24a24a43a2b4a2ca4a23b2a24a24a43a2b4a2(e2axb)a4a23b2a24a24a43a2b4a2J4a43a2b2a2(3bce2axb2a)Γ(4a43a2b2a2+1)a4a43a2b2a2a4a23b2a22a2)}} Maple : cpu = 0.253 (sec), leaf count = 256

{y(x)=1(eax3(Y12a(4a23b2a)(eax32ac)_C1+J12a(4a23b2a)(eax32ac))c(Y12a4a23b(eax32ac)_C1+J12a4a23b(eax32ac))(4a23b+2a))(3Y1/24a23ba(1/23ceaxa)_C1+3J1/24a23ba(1/23ceaxa))1}

Hand solution

y=12b+12ce2ax+2ay+32y2

This is of the form y=f0+f1y+f2y2 with f0=12b+12ce2ax,f1=2a,f3=32. Hence it is Riccati non-linear first order. Transforming to second order ODE using y=uuf2=23uu

Hence y=23(uu(u)2u2) and equating this to RHS of the ODE gives

23(uu(u)2u2)=12b+12ce2ax+2a(23uu)+32(23uu)223uu+23(u)2u2=12b+12ce2ax43auu+23(u)2u223uu=12b+12ce2ax43auuuu=34b34ce2ax+2auuu=(34b+34ce2ax)u+2auu2au+34(b+ce2ax)u=0

This is second order linear ODE with varying coefficient. Solved using power series method giving solutions using special functions (Bessel functions). Let A=4a23ba,B=3ceaxa then

u(x)=C1eaxBesselJ(124a23ba,123ceaxa)+C2eaxBesselY(124a23ba,123ceaxa)

But

u(x)=C1aexp(ax)BesselJ(1/24a23ba,1/23cexp(ax)a)1/2C1exp(ax)(BesselJ(1/24a23ba+1,1/23cexp(ax)a)1/334a23bcexp(ax)BesselJ(1/24a23ba,1/23ceaxa))3cexp(ax)+C2aexp(ax)BesselY(1/24a23ba,1/23cexp(ax)a)1/2C1exp(ax)(BesselY(1/24a23ba+1,1/23cexp(ax)a)1/334a23bcexp(ax)BesselY(1/24a23ba,1/23ceaxa))3ceax

Hence from y=23uu the solution is now found.

Verification

ode:=2*diff(y(x),x)-3*y(x)^2-4*a*y(x)=b+c*exp(-2*a*x); 
uode:=diff(u(x),x$2)-2*a*diff(u(x),x)+3/4*(b+c*(exp(-2*a*x)))*u(x)=0; 
uSol:=dsolve(uode,u(x)); 
my_sol:=(-2/3)*diff(rhs(uSol),x)/rhs(uSol); 
odetest(y(x)=my_sol,ode); 
0