Finding the Solution to a Differential Equation Passing Through (1,2)

[Cadbury]

Hello!

How do you find the solution of this equation that passes through (1,2)?

2 xy dy/dx = y^2 - 2 (x^3)

I have a problem with using the latex feature sorry if it is hard to read :(
Thank you! :)

 

[evinda]

Hello!

How do you find the solution of this equation that passes through (1,2)?

2 xy dy/dx = y^2 - 2 (x^3)

I have a problem with using the latex feature sorry if it is hard to read :(
Thank you! :)

(Wave)

$$2xydy=(y^2-2x^3)dx \Rightarrow (y^2-2x^3)dx -2xydy=0$$

$$M=y^2-2x^3$$

$$\frac{\partial{M}}{\partial{y}}=2y$$

$$N=y^2-2x^3$$

$$\frac{\partial{N}}{\partial{x}}=-6x^2$$Can you compute now $\frac{\frac{\partial{M}}{\partial{y}}-\frac{\partial{N}}{\partial{x}}}{N}$ ?

 

[Cadbury]

(Wave)

$$2xydy=(y^2-2x^3)dx \Rightarrow (y^2-2x^3)dx -2xydy=0$$

$$M=y^2-2x^3$$

$$\frac{\partial{M}}{\partial{y}}=2y$$

$$N=y^2-2x^3$$

$$\frac{\partial{N}}{\partial{x}}=-6x^2$$Can you compute now $\frac{\frac{\partial{M}}{\partial{y}}-\frac{\partial{N}}{\partial{x}}}{N}$ ?

Hi! Thank you for your fast response hehe.
So,
= [ 2y+6(x^2) ] / [ (y^2) - 2 (x^3)] is this the answer?
because i thought i should be using x=u+h, y=v+k, the equation will then be homogenous and then seperable

 

[evinda]

Hi! Thank you for your fast response hehe.
So,
= [ 2y+6(x^2) ] / [ (y^2) - 2 (x^3)] is this the answer?
because i thought i should be using x=u+h, y=v+k, the equation will then be homogenous and then seperable

I accidentally took the same $M$ and $N$, I am sorry... (Blush)

So, it is like that:$$M=y^2-2x^3$$

$$N=-2xy$$

$$\frac{\partial{M}}{\partial{y}}=2y$$

$$\frac{\partial{N}}{\partial{x}}=-2y$$

$$\frac{\frac{\partial{M}}{\partial{y}}-\frac{\partial{N}}{\partial{x}}}{N}=-\frac{2}{x}$$

$$\mu(x,y)=e^{\int -\frac{2}{x}dx}=e^{-2 \ln |x|}=\frac{1}{x^2} $$Now we multilpy our equation by $\mu$ and we get:

$$(2x-\frac{y^2}{x^2})dx+\frac{2y}{x} dy =0 $$$$M'=2x-\frac{y^2}{x^2}\\N'=\frac{2y}{x}$$

$$\frac{\partial{M'}}{\partial{y}}=-\frac{2y}{x^2}$$

$$\frac{\partial{N'}}{\partial{x}}=\frac{-2y}{x^2}$$

So now $\frac{\partial{M'}}{\partial{y}}=\frac{\partial{N'}}{\partial{x}}$.Now you solve the equations $\frac{\partial F}{\partial y}=M', \frac{\partial F}{\partial x}=N' $.

 

[Euge]

Hello!

How do you find the solution of this equation that passes through (1,2)?

2 xy dy/dx = y^2 - 2 (x^3)

I have a problem with using the latex feature sorry if it is hard to read :(
Thank you! :)

Hi Cadbury,

Just like in one your earlier problems, noticing that $(y^2)' = 2yy'$ you may substitute $u = y^2$ and reduce your equation to the linear ODE $xu' - u = -2x^3$. Using the initial condition $y(1) = 2$, we find $u(1) = 4$. So now you have the IVP

$$xu' - u = -2x^3,\quad u(1) = 4,$$

which you can solve with the integrating factor approach.