Exact diferencial equations ?

Question

i need to know if it is exact and solve it.

the equation is: (y^2 cosx - 3x^2 * y - 2x)dx + (2ysenx - x^3 + lny)dy = 0

y(1) = 1<--------initial value.

the book´s answer: y^2 senx - x^3 * y - x^2 + ylny - y = 0

its kind of dificult for me so i need help. thanks

Answer 1

you have a P(x,y) dx + Q(x,y) dy = 0...exact if P_y = Q_x....P_y = 2y cos x - 3x² while Q_x = 2y cos x - 3x²...thus exact...so integrate the 1st term---&gt; &quot;answer&quot; =y² sin x - y x^3 - x² + C(y), take a derivative with respect to y of this----&gt; 2y sin y - x^3 +C &#039;(y), set equal to Q---&gt; C&#039;(y) = ln y---&gt; C(y) = y ln y - y...get the &quot;answer&quot;

Answer 2

OK - the equation will be exact if the partial derivative with respect to y of (y^2 cosx - 3x^2 * y - 2x) is equal to the partial derivative with respect to x of (2ysenx - x^3 + lny) . To represent partial derivatives I will use the letter q instead of symbol that is normally used.

q (y^2 cosx - 3x^2 * y - 2x) / qy =

2ycosx - 3x^2 - 0

q ((2ysenx - x^3 + lny) /qx =

2ycosx - 3x^2 + 0

So now you know that the equation is exact. To solve the equation, note that

qf/qx = y^2 cosx - 3x^2 * y - 2x

Integrating both sides of this equation with respect to x produces

f(x,y) = y^2sinx - yx^3 - x^2 + g(y)

Where g(y) is the &quot;constant&quot; of integration. To figure out what g(y) is, you can differentiate both sides of the equation above with respect to y and set it equal to (2ysenx - x^3 + lny).

qf/qy = 2ysinx - x^3 + g&#039;(y) = (2ysenx - x^3 + lny)

g&#039;(y)  = lny

Integrating both sides via integration by parts you have

g(y) = ylny - y

And so the general solution is

y^2sinx - yx^3 - x^2 + g(y) = c

OR

y^2sinx - yx^3 - x^2 +  ylny - y = c

Umm... OK. I think I understand were you might be confused. Notice that the book&#039;s answer does not satisy the initial condition y(1)=1.

1*sin(1) - 1*1 - 1 - 1*ln1 - 1 ≠ 0

In other words, either you did not write the initial condition correctly or  the book&#039;s answer is wrong.