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is the answer key wrong for this exact equation?

find_the_fun

Active member
Feb 1, 2012
166
For the D.E. \(\displaystyle (x-y^3+y^2 \sin{x})dx = (3xy^2+2y \cos{x}) dy\) is the solution \(\displaystyle xy^3+\frac{x^2}{2}+y^2 \cos{x}=c\) ? That's what I got but the back of book said \(\displaystyle \frac{x^2}{2}\) should negative, not positive. How does this happen? I get \(\displaystyle \frac{x^2}{2}\) from integrating x so where does the negative come from?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The solution you give (and your book) does not go with the given ODE.
 

find_the_fun

Active member
Feb 1, 2012
166

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For the D.E. \(\displaystyle (x-y^3+y^2 \sin{x})dx = (3xy^2+2y \cos{x}) dy\) is the solution \(\displaystyle xy^3+\frac{x^2}{2}+y^2 \cos{x}=c\) ? That's what I got but the back of book said \(\displaystyle \frac{x^2}{2}\) should negative, not positive. How does this happen? I get \(\displaystyle \frac{x^2}{2}\) 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:

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

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

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

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

Differentiating with respect to $y$, we find:

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

Hence:

\(\displaystyle g'(y)=-3xy^2-2xy\)

Thus:

\(\displaystyle g(y)=-xy^3-xy^2\)

Hence, the solution is given implicitly by:

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

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

Which we may arrange as:

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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

[tex]\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*}[/tex]

and

[tex]\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*}[/tex]

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

find_the_fun

Active member
Feb 1, 2012
166
Well, let's see what we find. I would begin by writing the ODE in standard differential form:

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

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

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

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

Differentiating with respect to $y$, we find:

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

Hence:

\(\displaystyle g'(y)=-3xy^2-2xy\)

Thus:

\(\displaystyle g(y)=-xy^3-xy^2\)

Hence, the solution is given implicitly by:

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

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

Which we may arrange as:

\(\displaystyle xy^3-\frac{x^2}{2}+y^2\cos(x)=C\)
I see where I made the mistake.

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

Differentiating with respect to $y$, we find:
\(\displaystyle -3xy^2-2y \cos{x}=2xy-2y\cos(x)+g'(y)\)

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