Proof of theorem 135 if y f t y is homogeneous then

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Proof of Theorem 1.3.5 : If y 0 = f ( t, y ) is homogeneous, then we saw in one of the remarks above that the equation can be written as y 0 = F ( y/t ), where F ( y/t ) = f (1 , y/t ). Introduce the function v = y/t into the differential equation, y 0 = F ( v ) . We still need to replace y 0 in terms of v . This is done as follows, y ( t ) = t v ( t ) y 0 ( t ) = v ( t ) + t v 0 ( t ) . Introducing these expressions into the differential equation for y we get v + t v 0 = F ( v ) v 0 = ( F ( v ) - v ) t v 0 ( F ( v ) - v ) = 1 t . The equation on the far right is separable. This establishes the Theorem. Example 1.3.10 : Find all solutions y of the differential equation y 0 = t 2 + 3 y 2 2 ty . Solution: The equation is Euler homogeneous, since f ( ct, cy ) = c 2 t 2 + 3 c 2 y 2 2 ctcy = c 2 ( t 2 + 3 y 2 ) 2 c 2 ty = t 2 + 3 y 2 2 ty = f ( t, y ) .
G. NAGY – ODE January 13, 2015 27 The next step is to compute the function F . Since we got a c 2 in numerator and denominator, we choose to multiply the right-hand side of the differential equation by one in the form (1 /t 2 ) / (1 /t 2 ), y 0 = ( t 2 + 3 y 2 ) 2 ty 1 t 2 1 t 2 y 0 = 1 + 3 y t 2 2 y t . Now we introduce the change of unknown v = y/t , so y = t v and y 0 = v + t v 0 . Hence v + t v 0 = 1 + 3 v 2 2 v t v 0 = 1 + 3 v 2 2 v - v = 1 + 3 v 2 - 2 v 2 2 v We obtain the separable equation v 0 = 1 t 1 + v 2 2 v . We rewrite and integrate it, 2 v 1 + v 2 v 0 = 1 t Z 2 v 1 + v 2 v 0 dt = Z 1 t dt + c 0 . The substitution u = 1 + v 2 ( t ) implies du = 2 v ( t ) v 0 ( t ) dt , so Z du u = Z dt t + c 0 ln( u ) = ln( t ) + c 0 u = e ln( t )+ c 0 . But u = e ln( t ) e c 0 , so denoting c 1 = e c 0 , then u = c 1 t . Hence, the explicit form of the solution can be computed as follows, 1 + v 2 = c 1 t 1 + y t 2 = c 1 t y ( t ) = ± t c 1 t - 1 . C Example 1.3.11 : Find all solutions y of the differential equation y 0 = t ( y + 1) + ( y + 1) 2 t 2 . Solution: This equation is not homogeneous in the unknown y and variable t , however, it becomes homogeneous in the unknown u ( t ) = y ( t ) + 1 and the same variable t . Indeed, u 0 = y 0 , thus we obtain y 0 = t ( y + 1) + ( y + 1) 2 t 2 u 0 = tu + u 2 t 2 u 0 = u t + u t 2 . Therefore, we introduce the new variable v = u/t , which satisfies u = t v and u 0 = v + t v 0 . The differential equation for v is v + t v 0 = v + v 2 t v 0 = v 2 Z v 0 v 2 dt = Z 1 t dt + c, with c R . The substitution w = v ( t ) implies dw = v 0 dt , so Z w - 2 dw = Z 1 t dt + c - w - 1 = ln( | t | ) + c w = - 1 ln( | t | ) + c . Substituting back v , u and y , we obtain w = v ( t ) = u ( t ) /t = [ y ( t ) + 1] /t , so y + 1 t = - 1 ln( | t | ) + c y ( t ) = - t ln( | t | ) + c - 1 . C
28 G. NAGY – ODE january 13, 2015 1.3.3. Exercises. 1.3.1.- Find all solutions y to the ODE y 0 = t 2 y . Express the solutions in explicit form. 1.3.2.- Find every solution y of the ODE 3 t 2 + 4 y 3 y 0 - 1 + y 0 = 0 . Leave the solution in implicit form. 1.3.3.- Find the solution y to the IVP y 0 = t 2 y 2 , y (0) = 1 . 1.3.4.- Find every solution y of the ODE ty + p 1 + t 2 y 0 = 0 . 1.3.5.- Find every solution y of the Euler homogeneous equation y 0 = y + t t . 1.3.6.- Find all solutions y to the ODE y 0 = t 2 + y 2 ty . 1.3.7.- Find the explicit solution to the IVP ( t 2 + 2 ty ) y 0 = y 2 , y (1) = 1 . 1.3.8.- Prove that if y 0 = f ( t, y ) is an Euler homogeneous equation and y 1 ( t ) is a so- lution, then y ( t ) = (1 /k ) y 1 ( kt ) is also a solution for every non-zero k R .
G. NAGY – ODE January 13, 2015 29 1.4. Exact Equations A differential equation is called exact when it can be written as a total derivative of an appropriate function, called potential function. When the equation is written in that way it

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