Question

Find the value of b for which the given equation is exact then solve it using that value of b.

\((ye^{2xy}+x)+bxe^{2xy}y\prime=0\)

Answer 1

My book says that b=1, but when I try and show that it is exact.... I am unable to prove this.
Any thoughts?

Answer 2

Thought of any conditions yet? A differential equation is considered as exact if and only if:
\[\Large M_y=N_x \]
so in your case that would mean:

\[\Large e^{2xy}+2xye^{2xy}=be^{2xy}+2bxye^{2xy} \]
if b=1 then the statement is correct. you could also simplify the expression a bit of course to get a better overview.

Answer 3

So, basically it has to be this.... if b=1

\(\dfrac{\partial M}{dy}(ye^{2xy}+x)=\dfrac{\partial N}{dx}(xe^{2xy})\)

Answer 4

But when I try evaluating I get different answers.
(屮ﾟДﾟ)屮

Answer 5

\[\Large \frac{\partial}{\partial y}(ye^{2xy}+x)=e^{2xy}+2xye^{2xy} \] correct=

Answer 6

When I take the derivative, I get.
\(\Large{e^{2xy}}\)

Answer 7

you see that you need to apply the product rule to this? since you have another y in the exponent.

Answer 8

If that confuses you, just remember what the partial differential means:
\[\Large \frac{\partial }{\partial y} \]means that you keep \(x\) constant, so you can set \(x=c\) and then derive as if it were a normal single variable expression:
\[\Large \frac{d}{dy}(ye^{2cy}) =e^{2cy}+2cye^{2cy}\] you will see that you need the product rule here, now you can back substitute \(c=x\)