- #1
- 3,264
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$ 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
B.2.2.2 solve DE ....separate variables
- MHB
- Thread starter karush
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- Variables
So, what Romsek is saying is that you can "multiply" both sides by dx, and then use the properties of derivatives to get y'= y(x+h) and dy= f'(x)dx.f
$ 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
- #2
- 30
- 0
\(\displaystyle y^\prime = \dfrac{x^2}{y(1+x^3)}\\~\\
y dy = \dfrac{x^2}{1+x^3}dx\\~\\
\dfrac 1 2 y^2 = \dfrac{1}{3} \log \left(x^3+1\right) + C\\~\\
y^2 = \dfrac 2 3 \log(x^3+1) + C~\text{(note the constant absorbs multiplication by other constants)}\\~\\
y = \pm \left(\dfrac 2 3 \log(x^3+1) + C\right)^{1/2} = \pm \sqrt{\dfrac 2 3} \left(\log(x^3+1)+C\right)^{1/2}\\~\\
\text{$C$ is in the answer because there is a whole family of curves parameterized by $C$ that are a solution to this diff eq.}\\~\\
\text{You can select a specific curve by specifying initial conditions.}
\)
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.}
\)
- #3
- 3,264
- 4
ok i don't see how you can just
put dx in separations
great help tho appreciate the steps
put dx in separations
great help tho appreciate the steps
- #4
- 1,948
- 976
- #5
- 3,264
- 4
the second step the dy was already there the dx wasn't
- #6
- 30
- 0
\(\displaystyle y^\prime = \dfrac{dy}{dx}\)
so just "multiply" both sides by \(\displaystyle dx\)
Have you read on how to solve separable diff eqs?
so just "multiply" both sides by \(\displaystyle dx\)
Have you read on how to solve separable diff eqs?
- #7
- 44
- 0
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.
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.