# Math 53: Solutions to troublesome midterm problems

5) Making the substitution u=x+y, we obtain

du/dx=1+1/u.

We now try separation of variables. Dividing both sides by 1+1/u, and integrating, we obtain

u-log|u+1|=x+c.

The initial condition says u=-1 when x=0. But log|u+1| is not defined if u=-1. Does this mean the equation has no solution?

No! When we obtained the above family of solutions, we divided by 1+1/u, which means we assumed that 1+1/u is not zero. But this assumption is not always valid. There is an additional solution with 1+1/u=0, namely

u=-1.

This solution does satisfy the initial condition, and it translates into

y=-x-1.

Alternate solution: think of x as a function of y and turn both fractions upside down. Then you get a linear equation for x which you can easily solve. (I didn't teach you this method but it is sometimes useful.)

6b) True. It is straightforward to check that L_1+L_2 is linear, i.e. preserves addition and scalar multiplication, using the fact that L_1 and L_2 are linear. This is similar to the homework problem showing that the composition of two linear operators is linear.

6c) False. dy/dx=y^{38}-1 is positive if y=2. If we start at y=2, it is impossible to go below y=2 (let alone converge to y=1), because any such trajectory would, at some time x, have to go down at y=2, which is impossible since dy/dx is positive when y=2, regardless of x. (For autonomous equations, what goes up must not come down.)

Alternate explanation: y=1 is an unstable critical point, because d/dy(y^{38}-1)=38y^{37}=38>0. It is impossible to converge to an unstable equilibrium unless you start there.

6d) True. By the existence theorem, there is a solution satisfying the three initial conditions y(5)=1, y'(5)=2, and y''(5)=3. We cannot choose the value of y'''(5), but y'''(5) must equal something. Putting x=5 into the differential equation and using the above three conditions, we see that y'''(5)=4. This is what we wanted, so a solution exists.

For four random initial conditions, a solution would probably not exist, but we got "lucky". This is related to the fact that if you have four equations for three unknowns, a solution might exist, namely when one of the equations is "redundant".