MHB Finding D.E Solutions for $x\sin(y)+x^2y=c$ and $3x^2-xy^2=c$

[bergausstein]

FIND THE D.E DESIRED.

a. $x\sin(y)+x^2y=c$

here i obtain the first derivative,

$\displaystyle x\cos(y)+\sin(y)+x^2yy'+2xy=c$

b. $3x^2-xy^2=c$

here I also obtain the derivative of the eqn.

$6x-x^2xyy'-y^2=c$can you help continue with these problems? thanks!

 

[MarkFL]

Re: finding D.E desired.

Let's work the first problem. We are given:

$$x\sin(y)+x^2y=c$$

Now, what we want to do is implicitly differentiate with respect to $x$, and bear in mind that $y$ is a presumably a function of $x$, so we must use the chain rule when we differentiate.

Also recall that the derivative of a constant is zero. Your application of the chain rule is incorrect on the first product on the left, and your application of the product rule is incorrect on the second product. Look over what you did and see if you can spot the errors (both of which are minor) and correct them. :D

 

[bergausstein]

Re: finding D.E desired.

differentiating with respect to x

$\displaystyle xy'\cos(y)+\sin(y)+x^2y'+2xy=c$

what's next?

 

[MarkFL]

Re: finding D.E desired.

differentiating with respect to x

$\displaystyle xy'\cos(y)+\sin(y)+x^2y'+2xy=c$

what's next?

You actually want:

$$xy'\cos(y)+\sin(y)+x^2y'+2xy=0$$

Recall that $c$ is a constant, and as such its derivative is zero. Now what you do next depends on whether you want the ODE in the form:

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

or:

$$M(x,y)\,dx+N(x,y)\,dy=0$$

If you want it in the first form, then simply solve for $y'$. I suspect this form will be fine.

 

[bergausstein]

Re: finding D.E desired.

my answer would be in the form

$\displaystyle (x \cos(y)+x^2)dy+ ( \sin(y)+2xy)dx=0$

how about the next problem?

this is what I have for 2nd prob

$\displaystyle y'=\frac{3}{y}-y$

or

$\displaystyle (6x-y^2)dx-2xydy=0$

 

[MarkFL]

Re: finding D.E desired.

...
how about the next problem?

Did you look over what you posted and see where your errors are? Try applying the product rule again in the second term on the left in the original equation, and use the fact that the derivative of a constant is zero. What do you find?

 

[bergausstein]

Re: finding D.E desired.

yes I already corrected that and here's what I get

$\displaystyle y'=\frac{3}{y}-y$

or

$\displaystyle (6x-y^2)dx-2xydy=0$

are they correct?

 

[MarkFL]

Re: finding D.E desired.

yes I already corrected that and here's what I get

$\displaystyle y'=\frac{3}{y}-y$

or

$\displaystyle (6x-y^2)dx-2xydy=0$

are they correct?

The second form is correct, but you made an error in the first. I presume you began with:

$$y'=\frac{6x-y^2}{2xy}$$ ?

 

[bergausstein]

Isn't valid? I just solved for y'. can you explain why?

 

[MarkFL]

Isn't valid? I just solved for y'. can you explain why?

Can you say that:

$$\frac{6x-y^2}{2xy}=\frac{3}{y}-y$$

is an identity?