Question

Let \(\alpha\) and \(\beta\) be real numbers such that the differential equation

Hint:

For exact differential equations of the form \(Mdx+Ndy=0\), always use the condition \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\).

Answer 1

Correct Answer: 8

Step 1: Name the parts.
Write the equation as $M\,dx+N\,dy=0$ with $M=y^3+\alpha x y^4-5x+\cos 2y$ and $N=3xy^2+20x^2y^3+\beta x\sin 2y$.

Step 2: Use the exact condition.
For the equation to be exact we need $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$.

Step 3: Differentiate $M$ in $y$.
\[ \frac{\partial M}{\partial y}=3y^2+4\alpha x y^3-2\sin 2y \]

Step 4: Differentiate $N$ in $x$.
\[ \frac{\partial N}{\partial x}=3y^2+40 x y^3+\beta\sin 2y \]

Step 5: Match like terms.
Comparing the $xy^3$ terms gives $4\alpha=40$, so $\alpha=10$. Comparing the $\sin 2y$ terms gives $\beta=-2$.

Step 6: Add them.
\[ \alpha+\beta=10-2=8 \]
\[ \boxed{8.0} \]