Proof of Theorem1.3.5:Ify0=f(t, y) is homogeneous, then we saw in one of theremarks above that the equation can be written asy0=F(y/t), whereF(y/t) =f(1, y/t).Introduce the functionv=y/tinto the differential equation,y0=F(v).We still need to replacey0in terms ofv. This is done as follows,y(t) =t v(t)⇒y0(t) =v(t) +t v0(t).Introducing these expressions into the differential equation forywe getv+t v0=F(v)⇒v0=(F(v)-v)t⇒v0(F(v)-v)=1t.The equation on the far right is separable. This establishes the Theorem.Example1.3.10:Find all solutionsyof the differential equationy0=t2+ 3y22ty.Solution:The equation is Euler homogeneous, sincef(ct, cy) =c2t2+ 3c2y22ctcy=c2(t2+ 3y2)2c2ty=t2+ 3y22ty=f(t, y).
G. NAGY – ODE January 13, 201527The next step is to compute the functionF. Since we got ac2in numerator and denominator,we choose to multiply the right-hand side of the differential equation by one in the form(1/t2)/(1/t2),y0=(t2+ 3y2)2ty1t21t2⇒y0=1 + 3yt22yt.Now we introduce the change of unknownv=y/t, soy=t vandy0=v+t v0. Hencev+t v0=1 + 3v22v⇒t v0=1 + 3v22v-v=1 + 3v2-2v22vWe obtain the separable equationv0=1t1 +v22v. We rewrite and integrate it,2v1 +v2v0=1t⇒Z2v1 +v2v0dt=Z1tdt+c0.The substitutionu= 1 +v2(t) impliesdu= 2v(t)v0(t)dt, soZduu=Zdtt+c0⇒ln(u) = ln(t) +c0⇒u=eln(t)+c0.Butu=eln(t)ec0, so denotingc1=ec0, thenu=c1t. Hence, the explicit form of the solutioncan be computed as follows,1 +v2=c1t⇒1 +yt2=c1t⇒y(t) =±t√c1t-1.CExample1.3.11:Find all solutionsyof the differential equationy0=t(y+ 1) + (y+ 1)2t2.Solution:This equation is not homogeneous in the unknownyand variablet, however,it becomes homogeneous in the unknownu(t) =y(t) + 1 and the same variablet. Indeed,u0=y0, thus we obtainy0=t(y+ 1) + (y+ 1)2t2⇔u0=tu+u2t2⇔u0=ut+ut2.Therefore, we introduce the new variablev=u/t, which satisfiesu=t vandu0=v+t v0.The differential equation forvisv+t v0=v+v2⇔t v0=v2⇔Zv0v2dt=Z1tdt+c,withc∈R. The substitutionw=v(t) impliesdw=v0dt, soZw-2dw=Z1tdt+c⇔-w-1= ln(|t|) +c⇔w=-1ln(|t|) +c.Substituting backv,uandy, we obtainw=v(t) =u(t)/t= [y(t) + 1]/t, soy+ 1t=-1ln(|t|) +c⇔y(t) =-tln(|t|) +c-1.C
28G. NAGY – ODEjanuary 13, 20151.3.3.Exercises.1.3.1.-Find all solutionsyto the ODEy0=t2y.Express the solutions in explicit form.1.3.2.-Find every solutionyof the ODE3t2+ 4y3y0-1 +y0= 0.Leave the solution in implicit form.1.3.3.-Find the solutionyto the IVPy0=t2y2,y(0) = 220.127.116.11.-Find every solutionyof the ODEty+p1 +t2y0= 0.1.3.5.-Find every solutionyof the Eulerhomogeneous equationy0=y+tt.1.3.6.-Find all solutionsyto the ODEy0=t2+y2ty.1.3.7.-Find the explicit solution to the IVP(t2+ 2ty)y0=y2,y(1) = 18.104.22.168.-Prove that ify0=f(t, y) is an Eulerhomogeneous equation andy1(t) is a so-lution, theny(t) = (1/k)y1(kt) is also asolution for every non-zerok∈R.
G. NAGY – ODE January 13, 2015291.4.Exact EquationsA differential equation is called exact when it can be written as a total derivative of anappropriate function, called potential function. When the equation is written in that way it