Businesses and Ever

Separation of Variables / Mixing Problems

1. Find the general solution of the differential equation dM = 2.4 − .2M . (Such a differential equation dt came up, for instance, when we modeled the amount of medicine in a patient’s body.) Solution. We can use separation of variables: − Integrating both sides, − Multiplying both sides by −0.2 gives ln |M − 12| = −0.2t − 0.2C. Since −0.2C is just an arbitray constant, we can give it a new name; let’s call it A. So, ln |M − 12| = A − 0.02t. Then, Again, ±eA is just an arbitrary constant, so let’s call it B. So, M − 12 = Be−0.02t , and M (t) = 12 + Be−0.02t .2

t 2. Last time, we solved the differential equation dy = − y by drawing the slope field, guessing the solution, dt and checking it. Now, solve the differential equation using separation of variables. 1 2.4−0.2M

dM = dt.1 Simplifying,

1 1 · dM = dt. 0.2 M − 12

1 ln |M − 12| = t + C. 0.2

M − 12 = ±eA e−0.02t .

Solution. We can rewrite Integrating both sides,

dy dt

t = − y as

y dy = −t dt. 1 2 1 y = − t2 + C. 2 2 y 2 = −t2 + 2C.

Multiplying both sides by 2, Since 2C is still just an arbitrary constant, we can give it a new name; let’s call it A. So, y 2 = A − t2 , √ and y = ± A − t2 .

1 Technically, we can only do this if 2.4 − 0.2M = 0; if 2.4 − 0.2M = 0, which happens when M = 12, the original differential equation is just dM = 0, so M (t) = 12 is a solution. dt 2 Since B = ±eA , B should technically be non-zero. But we remarked earlier that M (t) = 12 is a solution, so B = 0 is also okay.

1

3. Solve the differential equation tion y(0) = 1. Solution. We can rewrite

dy dt

dy dt

= e−t−y , and find the particular solution satisfying the initial condi-

= e−t−y as ey dy = e−t dt.

Integrating both sides, ey = −e−t + C. So, y(t) = ln(C − e−t ). Plugging in the initial condition gives 1 = ln(C − 1), so e = C − 1, and C = 1 + e. So, our answer is y(t) = ln(1 + e − e−t ) . 4. Solve the differential equation y ′ = 2y − 6. Solution. First, y(t) = 3 is a solution. If y = 3,...