# Math Help - implicit form

1. ## implicit form

Hi everyone,

If anyone could help to check if no mistake in my calculation.

Here is the question and my answer:

Hence find, an implicit form, the general solution of the differential equation

dy/dx = 2y3/2 cos x
(2 + sin x)2
The differential equation dy/dx = 2y3/2 cos x/(2+sin x)2,

Has a right-hand side of the form f(x) g(y), where

g(y) = 2y 3/2 cos x and f(x) = (2+sin x)2

Dividing both sides of the differential equation by g(y) means multiplying both sides by y. This give

∫ 2y 3/2 cos x dy = ∫(2+sin x)2 dx
y5/2-sin x + a = cos x3 + b

Thus the general solution, in implicit form, is

y 5/2 = cos x3 + c

2. Originally Posted by valerie-petit
Hi everyone,

If anyone could help to check if no mistake in my calculation.

Here is the question and my answer:

Hence find, an implicit form, the general solution of the differential equation

dy/dx = 2y3/2 cos x
(2 + sin x)2
The differential equation dy/dx = 2y3/2 cos x/(2+sin x)2,

Has a right-hand side of the form f(x) g(y), where

g(y) = 2y 3/2 cos x and f(x) = (2+sin x)2

Dividing both sides of the differential equation by g(y) means multiplying both sides by y. This give

∫ 2y 3/2 cos x dy = ∫(2+sin x)2 dx
y5/2-sin x + a = cos x3 + b

Thus the general solution, in implicit form, is

y 5/2 = cos x3 + c

Thse are just seperable differential equations

All you need to do is this

it will be of the form

$\frac{dy}{dx}=f(x)g(y)$

So you need to seperate to make it look like

$g(y)dy=f(x)dx$

Which implies that

$\int{g(y)}dy=\int{f(x)dx}$

Now let $G(y)$ and $F(x)$ be anti-derivatives respectively.

Then we have that

$G(y)=F(x)+C$

This implies that

$y=G^{-1}\left(F(x)+C\right)$

Where $G^{-1}$ denotes the inverse function of $G(y)$

3. Hello, Valerie!

If I read your typing correctly, you went way off . . .

Find, in implicit form, the general solution of the differential equation:

. . $\frac{dy}{dx} \;=\; \frac{2y^{\frac{3}{2}}\cos x}{(2 + \sin x)^2}$
Separate variables ... divide by $y^{\frac{3}{2}}$

. . $y^{-\frac{3}{2}} \;=\;\frac{2\cos x\,dx}{(2 + \sin x)^2}$

Integrate: . $-2y^{-\frac{1}{2}} \;=\;-2(2+\sin x)^{-1} + C \quad\Rightarrow\quad y^{-\frac{1}{2}} \;=\;(2+\sin x)^{-1} + C$

5. ## Explicit form

On this differential equation I was asked to find a particular solution with y= 4 and x= 0. Which I did, now I have to find the explicit form of that particular solution.

Here is the question and my answer:

Find the explicit form of this particular solution.
y-1/2 = (2+sin x)-1 + 3/8

The explicit solution is obtained by making y the subject of the equation

y= (2 + sin x)^1/2 +3/8

I multiplied -1/2 both side