How do you solve the differential equation: dy/dx= (2x^4+4x^2+8) / (x^3+4x)?

Answer 1

dy/dx = (2x⁴ + 4x² +8)/(x³ + 4x) = 2x + (8 - 4x²)/(x³ + 4x). Write (8 - 4x²)/(x(x² + 4)) = A/x + (Bx +C)/(x² + 4) and Ax² + 4A +Bx² +Cx = 8 - 4x². Then 8 = 4A or A = 2 and C = 0; A + B = -4 or B = -6.

Thus, dy/dx = 2x + 2/x - 6x/(x² + 4) which integrates to y = x² +2ln |x| - 3ln (x² + 4) + C.

Rewrite as y = x² + ln(x²/(x² + 4)³) + C.

Answer 2

1. Separate the variables by multiplying both sides by dx
2. Integrate

To do this integration I would first factor out a 2 from the numerator and put it outside the integral giving you:

∫dy = 2∫ (x⁴ + 2x² + 4)/(x³ + 4x))dx

Then divide the denominator into the numerator. Doing so gives you "x" with a remainder of "-2x² + 4" so the differential equation becomes:

∫dy = 2∫ (x + (-2x² + 4)/(x³ + 4x))dx

Split the right side into two parts like this:

∫dy = 2∫ x dx + 2∫ (-2x² + 4)/(x³ + 4x))dx

Integrating the left side and the left portion of the right side gives you:

y = 2(1/2x²) + 2∫ (-2x² + 4)/(x³ + 4x))dx

Factor out a -2 from the numerator of the integral as well as factor the denominator to make it:

y = 2(1/2x²) - 4∫ (x² - 4)/[(x)(x² + 4)]dx

Use Partial Fractions to separate the integral into:

y = 2(1/2x²) - 4[∫ -1/x dx + ∫2x/(x² + 4) dx ]

Then the integrals become log expressions:

y = 2(1/2x²) - 4[-ln(x) + 2ln(x² + 4)]

Simplify:

y = x² + 4ln(x) - 4ln(x² + 4) + C