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Example solving non-linear first order ODE

Nasser M. Abbasi

September 8, 2018 compiled on — Saturday September 08, 2018 at 05:23 AM

dy dt + y3 2 t = a3 2 y 0 = 0

Write as

y3 2 a3 2 dt + dy = 0 M t,ydt + N t,ydy = 0 (1)

Where

M = y3 2 a3 2 N = 1

Check if exact

M t,y y = 3 2y1 2 N t,y t = 0

Since M t,y y N t,y t then Not exact. Trying integrating factor A = N t M y M = 3 2 y 1 2 y 3 2 a 3 2 , Since it is a function of y alone, then it (1) can be made exact. The integrating factor is

μ = eAdy = e 3 2 y 1 2 y 3 2 a 3 2 dy = e lna 3 2 y 3 2 = 1 a3 2 y3 2

Multiplying (1) by this integrating factor, now it becomes exact

μM t,ydt + μN t,ydy = 0

Now we follow standard method for solving exact ODE. Let

dU dt = μM = y3 2 a3 2 a3 2 y3 2 = 1 (2) dU dy = μN = 1 a3 2 y3 2 (3)

From (2)

U = dt = t + f y (4)

Substituting this into (3) to solve for f y

fy = 1 a3 2 y3 2 f y = 2 3 3a arctan 1 + 2y a 3 2 3aln a y + 1 3aln a + ay + y + C

Hence the solution from (4) is

U = t + 2 3 3a arctan 1 + 2y a 3 2 3aln a y + 1 3aln a + ay + y + C

But dU dt = 0, hence U = C1. Therefore, collecting constants into one, the solution is (implicit form)

t + 2 3 3aarctan 1 + 2y a 3 + 2 3aln a y 1 3aln a + ay + y = C

From initial conditions

2 3 3aarctan 1 3 + 2 3aln a 1 3aln a = C C = 2 3 3a π 6 + 2 3aln a 1 3aln a C = 2 3 3a π 6 C = π 3 9a

Hence final solution for y t in implicit form is

t + 2 3 3aarctan 1 + 2y a 3 + 2 3aln a y 1 3aln a + ay + y = π 3 9a 3t a + 2 3arctan a + 2y 3 a + 6ln a y ln a + ay + y = π 3 3