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: Identify \(M(x,y)\) and \(N(x,y)\).
\[ M=y^3+\alpha xy^4-5x+\cos 2y \] \[ N=3xy^2+20x^2y^3+\beta x\sin 2y \]

Step 2: Exactness condition.
For exact differential equation,
\[ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} \]

Step 3: Differentiate \(M\) with respect to \(y\).
\[ \frac{\partial M}{\partial y} = 3y^2+4\alpha xy^3-2\sin 2y \]

Step 4: Differentiate \(N\) with respect to \(x\).
\[ \frac{\partial N}{\partial x} = 3y^2+40xy^3+\beta \sin 2y \]

Step 5: Compare both expressions.
\[ 3y^2+4\alpha xy^3-2\sin 2y = 3y^2+40xy^3+\beta \sin 2y \]

Step 6: Compare coefficients.
Comparing coefficient of \(xy^3\),
\[ 4\alpha=40 \] \[ \alpha=10 \]
Comparing coefficient of \(\sin 2y\),
\[ \beta=-2 \]

Step 7: Find \(\alpha+\beta\).
\[ \alpha+\beta=10+(-2)=8 \] \[ \boxed{8.0} \]