How Can We Solve the Differential Equation dy/dx = 35/(y^(1/8) + 25x^2)?

[hardatwork]

Homework Statement

The differential equation dy/dx= 35/(y^1/8+25x²y^1/8 has an implicit general solution of the form F(x,y)=K. In fact, because the differential equation is separable, we can define the solution curve implicitly by the form F(x)=G(x)+H(y)=K.
Find such solution and then give the related functions requested.

Homework Equations

The Attempt at a Solution

dy/dx=35/(y^1/8(1+25x²)
1+25x²/35 dx=y^1/8dy
1/35$$\int1+25x^2 dx$$=$$\int y^(1/8) dy$$
x+25x³/105=8/9y^9/8
105(8/9y^9/8)-x-25³=K
so G(x)=x+25x³ and H(y)=8/9y^9/8

 

[hardatwork]

The problem is that this is incorrect and I don't know what I did wrong. Can someone see my mistake. Thank You in advance

 

[Subdot]

dy/dx=35/(y^1/8(1+25x²)
1+25x²/35 dx=y^1/8dy

The bolded line is where you went wrong. Look carefully at what you did. It may be easier to see if you wrote it out "properly:"

$$\frac{dy}{dx} = \frac{35}{y^{1/8} + y^{1/8}25x^2}$$

$$\frac{dy}{dx} = \frac{35}{y^{1/8}(1 + 25x^2)}$$

Edit: I'm assuming the bolded line really reads ((1+25x²)/35)dx=y^1/8dy

 

[hardatwork]

So when I separate the variables it should be
y^1/8dy=$$35/1+25x^2 dx$$

 

[Subdot]

Assuming you mean y^1/8dy=(35/(1+25x²))dx, then yes, that is correct. It is more easily read as (35dx)/(1+25x²) or 35dx/(1+25x²) though, imo.

 

[hardatwork]

Okay. Thank You