MHB Is the answer key wrong for this exact equation?

[find_the_fun]

For the D.E. [math](x-y^3+y^2 \sin{x})dx = (3xy^2+2y \cos{x}) dy[/math] is the solution [math]xy^3+\frac{x^2}{2}+y^2 \cos{x}=c[/math] ? That's what I got but the back of book said $$\frac{x^2}{2}$$ should negative, not positive. How does this happen? I get $$\frac{x^2}{2}$$ from integrating x so where does the negative come from?

 

[MarkFL]

The solution you give (and your book) does not go with the given ODE.

 

[find_the_fun]

The solution you give (and your book) does not go with the given ODE.

Ok fixed question.

 

[MarkFL]

For the D.E. [math](x-y^3+y^2 \sin{x})dx = (3xy^2+2y \cos{x}) dy[/math] is the solution [math]xy^3+\frac{x^2}{2}+y^2 \cos{x}=c[/math] ? That's what I got but the back of book said $$\frac{x^2}{2}$$ should negative, not positive. How does this happen? I get $$\frac{x^2}{2}$$ from integrating x so where does the negative come from?

Well, let's see what we find. I would begin by writing the ODE in standard differential form:

$$\left(x-y^3+y^2 \sin{x} \right)dx+\left(-3xy^2-2y \cos{x} \right)dy=0$$

We can see that the equation is exact, so we take:

$$F(x,y)=\int x-y^3+y^2 \sin{x}\,dx+g(y)$$

$$F(x,y)=\frac{x^2}{2}+xy^2-y^2\cos(x)+g(y)$$

Differentiating with respect to $y$, we find:

$$-3xy^2-2y \cos{x}=2xy-2y\cos(x)+g'(y)$$

Hence:

$$g'(y)=-3xy^2-2xy$$

Thus:

$$g(y)=-xy^3-xy^2$$

Hence, the solution is given implicitly by:

$$\frac{x^2}{2}+xy^2-y^2\cos(x)-xy^3-xy^2=C$$

$$\frac{x^2}{2}-y^2\cos(x)-xy^3=C$$

Which we may arrange as:

$$xy^3-\frac{x^2}{2}+y^2\cos(x)=C$$

 

[Prove It]

I personally prefer the method where we perform both integrations then get the final solution by comparing the parts from the previous solutions.

We have

\displaystyle \begin{align*} \frac{\partial F}{\partial x} &= x - y^3 + y^2\sin{(x)} \\ F &= \int{ x - y^3 + y^2\sin{(x)}\,dx} \\ F &= \frac{x^2}{2} - x\,y^3 - y^2\cos{(x)} + g(y) \end{align*}

and

\displaystyle \begin{align*} \frac{\partial F}{\partial y} &= -3x\,y^2 - 2y\cos{(x)} \\ F &= \int{ -3x\,y^2 - 2y\cos{(x)}\,dy} \\ F &= -x\,y^3 - y^2\cos{(x)} + h(x) \end{align*}

Comparing the two solutions, it's clear that the solution has to be \displaystyle \begin{align*} F(x, y) = \frac{x^2}{2} - x\,y^3 - y^2\cos{(x)} + C \end{align*}

 

[find_the_fun]

Well, let's see what we find. I would begin by writing the ODE in standard differential form:

$$\left(x-y^3+y^2 \sin{x} \right)dx+\left(-3xy^2-2y \cos{x} \right)dy=0$$

We can see that the equation is exact, so we take:

$$F(x,y)=\int x-y^3+y^2 \sin{x}\,dx+g(y)$$

$$F(x,y)=\frac{x^2}{2}+xy^2-y^2\cos(x)+g(y)$$

Differentiating with respect to $y$, we find:

$$-3xy^2-2y \cos{x}=2xy-2y\cos(x)+g'(y)$$

Hence:

$$g'(y)=-3xy^2-2xy$$

Thus:

$$g(y)=-xy^3-xy^2$$

Hence, the solution is given implicitly by:

$$\frac{x^2}{2}+xy^2-y^2\cos(x)-xy^3-xy^2=C$$

$$\frac{x^2}{2}-y^2\cos(x)-xy^3=C$$

Which we may arrange as:

$$xy^3-\frac{x^2}{2}+y^2\cos(x)=C$$

I see where I made the mistake.

Original equation: $$(x-y^3+y^2\sin{x})dx-(3xy^2+2y\cos{x})dy=0$$

Differentiating with respect to $y$, we find:

$$-3xy^2-2y \cos{x}=2xy-2y\cos(x)+g'(y)$$

I didn't think the minus sign expanded out and had [math]3xy^2+2y \cos{x}[/math] :/