I am having a problem with the (2).Please help me.
Text-book Question:
Consider the IVP:
$$\color{crimson}{(105\sin(3y-15x)-y)dx+(-21\sin(3y-15)-x+2y)dy=0}$$
$$\color{#06f}{y(8)=40}$$
(1) Verify that the ODE is exact and Find $u(x,y)$:
$$u(x,y)=7\cos(3y-15x)-yx+y^2$$
(2) Hence find the solution to the IVP, in implicit form:
Now since this question says Hence, and also my notes say that the solution for a IVP is $u(x,y)=C$, so this should mean that:
$$7\cos(3y-15x)-yx+y^2=C$$
So the only way to find C, is to use $\color{#06f}{y(8)=40}$, so I did that:
$$C=7\cos(0)-40(8)+40^2=1287$$
$$\color{grey}{\text{which makes no sense...}}$$
So I tried this:
$$\color{crimson}{(105\sin(3y-15x)-y)dx+(-21\sin(3y-15)-x+2y)dy=0}$$
$$(105\sin(3y-15x)-y)dx=(21\sin(3y-15)+x-2y)dy$$
$$-y^2+xy-7\cos(3y-15x)=7\cos(15x-3y)-xy$$
$$\color{green}{\text{Which gets me lost...}}$$
My question: What is the question actually asking me to do?
Just a $\color{red}{\mbox{typo}}$, I think, in the first place. The differential equation is an exact one if: $$ \left[105\sin(3y-15x)-y\right]dx+\left[-21\sin(3y-15\color{red}{x})-x+2y\right]dy=0 $$ So it is supposed that such is the case. When put: $$ du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy = 0 $$ then we have: $$ \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right) = \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right) = 315\cos(3 y - 15 x) - 1 $$ And the original ODE is reproduced with: $$ u(x,y)=7\cos(3y-15x)-yx+y^2 \quad \Longrightarrow \quad \left\{ \begin{array}{l} \partial u/\partial x = 105\sin(3 y - 15 x) - y \\ \partial u/\partial y = -21\sin(3 y - 15 x) - x + 2 y \end{array} \right. $$ We know that $du=0$ , so indeed $u(x,y) = C$ is the solution. However, $C$ can be anything, not just $C=1287$ ; I don't see either how to make sense of the additional "condition" $y(8)=40$ .