Integrating Separable Equations: Comparing Solutions to Practice Problems

[Chandasouk]

I have solutions for 2 problems but they are different from the ones my book provides. This may be due to some simplification they chose to do, but I am uncertain.

1) dy/dx = x²/y

ydy = x²dx

Integrate both sides and you get

y²/2 = x³/3 + C

My book gave 3y² - 2x³ = C

2) dy/dx = x² / y(1+x³)

y(1+x³)dy = x²dx

ydy = x²/(1+x³) dx

Integrating both sides, I got

y²/2 = 1/3*ln(1+x³) + C

but my book gave

3y²-2ln(1+x³) = C

 

[Dick]

C is arbitrary. It could be anything. 6*C is also arbitrary. You may as well just label 6*C as C. Multiply both of your answers by 6. If C is arbitrary then e^C, 6C, C-1, etc etc are also arbitrary. Just call them C.

 

[Chandasouk]

Oh, I see. So my answers were correct. It seems my book likes not having fractions. Thanks again.

 

[Dick]

Oh, I see. So my answers were correct. It seems my book likes not having fractions. Thanks again.

Right, there's not single right expression. The answer keys will usually pick the simplest form of the constant. You should try and do that too.