Math 201 Lecture 05 Bernoulli and Linear Coefficients

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1 Math 0 Lecture 05 Bernoulli and Linear Coefficients Jan. 8, 0 Many eamples here are taken from the tetbook. The first number in () refers to the problem number in the UA Custom edition, the second number in () refers to the problem number in the 8th edition. Homogeneous equations: with f(t,ty)=f(,y) for any t. 0. Review =f(,y) (). Check that the equation is indeed homogeneous; We can either use the definition: Calculate f(t, t y), simplify, and see whether it turns out to be f(,y); Or in many cases we can simply try to write everything in the ratio y/.. Replace y by v in to obtain = d(v) = +v. () 3. Replace every y in f(,y) by v. Simplify. All s should disappear and the result should be a function of v, G(v). (If after simplification there are still some dangling around, you may want to return to step and check whether the equation is homogenous after all) 4. The new equation is +v=g(v) =[G(v) v]. (3) It is separable. Solve it. (Don t forget to check etra solution!) 5. Get the final answer y by y=v(). Simplify. If the formula for v is implicit, replace every v in your formula by y/, simplify. Equations of the form:. Decide that the equation is indeed of the form. Let z=a+by. We have So the equation becomes =G(a+by) (4) =G(a+by). (5) y= z a b = dz b a b. (6) dz b a =G(z) (7) b which is separable. (Again, if after simplification the equation is just not separable, that may mean the original equation is not of the form =G(a+by). 3. Solve the separable equation (Don t forget to check etra solutions!). Use y= z a b to get the final answer. If the solution to the separable equation is implicit, replace z by a+by, simplify. (8)

2 Math 0 Lecture 05 Bernoulli and Linear Coefficients Quiz: Solve ( = y + y ), y()=. (9) Solution. First check the equation is homogeneous: f(,y)= y + y Now let v=y/. We have so the equation for v is Separate variables: Integrate t f(t,ty)= ty + ty t = y + y =f(,y). (0) = +v, y + y = +v () v +v= v +v = v. () v=. (3) v = ln v =ln +C. (4) Now as the v equation is separable, we should check whether any etra solution eists. Recall the v equation = (5) v we have p(v)=. There is no constant c making p(c)=0. So no etra solution. v Finally back to y. Replace v= y we have y =ln +C y = ( ln +C). (7) This is the general solution. To solve the initial value problem, use the initial condition: y()= 0 =,y 0 =. Substitute into the equation we have So the solution to the initial value problem is.. Bernoulli Equations. = ( ln+c) C=4. (8) y = ( ln +4). (9). Basic Information The Equation +P()y=Q()yn. (0) where n is a real number. It looks very similar to the linear equation ecept a factor y n on the right hand side. +P()y=Q() (). For those who care about eistence and uniqueness, note that for any given C, the solution y = ( ln +C). (6) is not defined for <e C/. However this does not contradict the eistence and uniqueness theorems. Try plot the solutions to understand why.

3 Jan. 8, 0 3 How to get general solution Procedure:. Divide both sides by y n :. Set The equation becomes or equivalently This is linear. 3. Solve the linear equation for v. y n +P()y n =Q(). () v=y n. (3) +P()v=Q() (4) n +( n)p()v=( n)q(). (5) 4. Get y through y=v /( n). Add etra solution y=0 if n>0. The above obviously doesn t work if n=. However in this case, the equation becomes =(Q() P())y (6) which is separable. Also note that the case n=0 is eactly the linear case. Eample: (.6 3;.6 3) Solve = y y. (7) Solution. First note that y=0 is a solution. So we need to add it back at the end. Now divide both sides by y and let v=y. We have y = y + v=. (8) Recalling the method of solving linear equations, we compute = ln e = (9) Thus multiply both sides by, we reach Integrating, we obtain 4 = +v= d ( v). (30) v= 5 5 +C v= 5 3 +C. (3) Summarizing, the solutions are y= 5 3 +C, y=0. (3) Finally check solution. Clearly y=0 solves the equation. We have [ ] 3 = 5 3 +C 5 C 3 ; (33) y = 5 3 +C [ ] = 5 3 +C 5 +C 3. (34)

4 4 Math 0 Lecture 05 Bernoulli and Linear Coefficients This gives So the solution is correct. How to solve initial value problem (IVP) Solve y = 5 3 +C = y. (35) = y y, y()=. (36) Solution. Take the general solution y= 5 3 +C, y=0. (37) Substitute y=,= into the formula (As clearly y=0 does not satisfy the initial condition) we reach = 5 +C C= 4 5. (38) So the solution is ( y= ) 5. (39) How to check solutions. Nothing new here... Equations with Linear Coefficients. The Equation with a,b,c,a,b,c constants. How to get general solution (a +b y+c )+(a +b y+c )=0. (40) Idea: If we can get rid of the constants c,c, the equation becomes homogeneous and can be solved. Procedure:. Find constants h, k such that This is a by linear system for h,k: a (u+h)+b (v+k)+c =a u+b v; (4) a (u+h)+b (v+k)+c =a u+b v. (4) a h+b k = c (43) a h+b k = c (44). Onceh,k are obtained, re-write the equation using the new unknown v and new variable u. For the particular h,k found above, the equation becomes which is homogeneous: (a u+b v)du+(a u+b v)=0 (45) du = a u+b v a u+b v = a +b (v/u) a +b (v/u). (46) 3. Solve this homogeneous equation (Don t forget possible etra solutions!) 4. Replace u= h,v=y k. Simplify.

5 Jan. 8, 0 5 Eample: (.6 3;.6 3) Solve Solution. Write Set We obtain ( y)+(4+y 3)=0. (47) y = 4+y 3. (48) =u+h, y=v+k. (49) u v+[h k] = du 4u+v+[4h+k 3]. (50) Thus we need to find h,k such that The solution is Net we solve As this is a homogeneous equation, set We have u dz du +z= z 4+z u dz +3z+ du = z 4+z h k = 0 (5) 4h+k 3 = 0. (5) h=/, k=. (53) du = u v 4u+v. (54) z= v u. (55) 4+z z +3z+ dz= du u. (56) As usual for separable equations, we check possible zeros of z +3z+. We find two constant 4+z solutions z=,z=. To integrate the LHS, we write Setting we have Thus 4+z z +3z+ = a z+ + b z+ = (a+b)z+(a+b) z. (57) +3z+ a+b=, a+b= 4 (58) a= 3, b=. (59) 4+z z +3z+ dz = 3 z+ dz+ z+ dz = d[ 3 ln z+ +ln z+ ]. (60) Consequently which gives 3 ln z+ +ln z+ =ln u +C (6) (z+) =e C z+ 3 u. (6) As C is arbitrary, e C is an arbitrary positive real number. So the above, together with the solution z=, is equivalent to (z+) =C(z+) 3 u (63)

6 6 Math 0 Lecture 05 Bernoulli and Linear Coefficients where C is an arbitrary real number. Recalling z= v, the above becomes u Now as v=y k=y, u= h= /, we have which is equivalent to (v+u) =C(v+u) 3. (64) (y+ ) =C(y+ 3/) 3 (65) (y+ ) =C(y+ 3) 3 (66) due to the fact that C is an arbitrary real number. Note that the above formula alrea includes the case z=. The case z= corresponds to v= u or equivalently y= 3. Summarizing, the solutions to our problem are y= 3, (y+ ) =C(y+ 3) 3. (67) We check the solutions. For y= 3, we have =. Now re-write the equation: y ( y)+(4+y 3)=0 = 4+y 3 When y= 3, the right hand side becomes y ( 3 = ) 3 4+y = =. (69) So the etra solution y= 3 is indeed as solution. For the general formula we write it as and take d: [ ] (y+ ) d (y+ 3) 3 (68) (y+ ) =C(y+ 3) 3, (70) (y+ ) =C (7) (y+ 3) 3 = (y+ )(+) (y+ 3) 3 3 (y+ ) (+) (y+ 3) 4 = (y+ ) (y+ 3) 4[(y+ 3)(+) 3(y+ )(+)] = (y+ ) (y+ 3) 4 [(y 4)+( y 8+6)] = (y+ ) (y+ 3) 4 [( y)+(4+y 3)]. (7) Now we are sure that our solution is correct. How to solve initial value problem (IVP) Solve Solution. The general solutions are ( y)+(4+y 3)=0, y()=. (73) y= 3, (y+ ) =C(y+ 3) 3. (74) First we check that y= 3 clearly does not satisfy the initial condition. Now take the formula (y+ ) =C(y+ 3) 3 (75)

7 Jan. 8, 0 7 and substitute =,y=, we have So the solution to the initial value problem is C=. (76) How to check solutions (y+ ) =(y+ 3) 3. (77) See Common Mistakes for eamples. Don t forget to return to the original and y.. Things to be Careful/Tricky Issues 3. More Eamples To solidify your knowledge of Chapter, it is a good idea to work on the Review Problems at the end of the Chapter. This is more rewarding than working on the problems after each section, as much difficulty in solving first order equations comes from the need to decide its type first. And doing eercises after each section cannot test/practise this ability. So far we have studied 7 types of equations: Separable, Linear, Eact, Homogeneous, No Name ( =G(a+b)), Bernoulli, Linear Coefficients. Make sure you know how to spot them. Make sure you know how to solve them. Hints on Review Problems 30 (The numberings in 7th and 8th editions are identical): Note. Some of the review problems are related to section.5, which we do not require in 0. For these I will note IF (Integrating factor).. Separable;. Linear; 3. Eact (Note that the equation is given in a tricky way, with before ); 4. Linear; 5. Eact; 6. Separable; 7. Separable; 8. Bernoulli; 9. Homogeneous; 0. Eact;. No Name;. IF; 3. Linear; 4. Linear; 5. No Name; 6. Linear; 7. Bernoulli; 8. No Name; 9. Homogeneous; 0. Bernoulli;. Linear Coeff;. Linear Coeff; 3. Linear coeff; 4. Eact; 5. IF; 6. Separable; 7. Linear Coeff; 8. Linear Coeff; 9. IF; 30. No Name. 4. Notes and Comments For equations with linear coefficients, there are overall four cases: We notice that When b =0, the equation is linear. When b =a, the equation is eact. When a b =a b, the equation is of the form =G(a+by). (78) Otherwise the equation can be solved by trying to transform it to a homogeneous equation. Note that the same approach also works for the more general case y a +b =f y+c. (79) a +b y+c. The case b =b =0 can also be made eact via multiplying appropriate µ(,y).