MHB B.2.2.2 solve DE ....separate variables

[karush]

Spoiler: book answers

#2

$ y'= \dfrac{x^2}{y(1+x^3)}$

Separate y dy =\dfrac{x^2}{(1+x^3)...

ok i tried to get the book ans but someahere derailed
why is =c in the answer

 

[romsek]

$$y^\prime = \dfrac{x^2}{y(1+x^3)}\\~\\

y dy = \dfrac{x^2}{1+x^3}dx\\~\\

\dfrac 1 2 y^2 = \dfrac{1}{3} \log \left(x^3+1\right) + C\\~\\

y^2 = \dfrac 2 3 \log(x^3+1) + C~\text{(note the constant absorbs multiplication by other constants)}\\~\\

y = \pm \left(\dfrac 2 3 \log(x^3+1) + C\right)^{1/2} = \pm \sqrt{\dfrac 2 3} \left(\log(x^3+1)+C\right)^{1/2}\\~\\

\text{$C$ is in the answer because there is a whole family of curves parameterized by $C$ that are a solution to this diff eq.}\\~\\

\text{You can select a specific curve by specifying initial conditions.}

$$

 

[karush]

ok i don't see how you can just
put dx in separations

great help tho appreciate the steps

 

[topsquark]

ok i don't see how you can just
put dx in separations

Huh? What step are you not understanding?

-Dan

 

[karush]

the second step the dy was already there the dx wasn't

 

[romsek]

$$y^\prime = \dfrac{dy}{dx}$$

so just "multiply" both sides by $$dx$$

Have you read on how to solve separable diff eqs?

 

[Kansas Boy]

Before you take Differential Equations you need to know Calculus really well! I think you need more practice in Calculus.

You say " the dy was already there the dx wasn't ".
Strictly speaking neither was in the original equation- it had y'. But I am sure you understand that y' and dy/dx are just different notations for the derivative of y with respect to x.

Romsek said "just multiply both sides by dx". Now, Strictly speaking "dy/dx" is NOT a fraction, it is rather the limit of fractions: $\lim_{h\to 0}\frac{y(x+h)- y(x)}{h}$. But it can be "treated like a fraction". To make use of that we define (typically in Calculus 3) the "differentials" dy and dx separately from dy/dx such that dy= y' dx. That allows us to treat dy/dx as if it were a fraction and say if dy/dx= f'(x) then dy= f'(x)dx.