The given differential equation is: \[ (7x^4 \cot y - e^x \csc y) \frac{dx}{dy} = x^5 \] First, we rearrange the equation to express \( \frac{dx}{dy} \): \[ \frac{dx}{dy} = \frac{x^5}{7x^4 \cot y - e^x \csc y} \] Now, let’s separate the variables. To do so, we’ll solve for \( \frac{dy}{dx} \) and then integrate: \[ \frac{dy}{dx} = \frac{7x^4 \cot y - e^x \csc y}{x^5} \] Now, evaluate the values at \( x = 1 \) and \( x = 2 \), and integrate accordingly to get \( y \).
We’re interested in \( \cos y \) at \( x = 2 \), so we need to evaluate the solution at this point.
After solving the equation and evaluating the expressions, we find that the correct value of \( \cos y \) at \( x = 2 \) is \( \frac{e^2}{128} - 1 \).
Thus, the correct answer is \( \frac{e^2}{128} - 1 \).
Step 1: The given differential equation is:
\[ \frac{dy}{dx} = \frac{7 \cot y}{x} - \frac{e^x \csc y}{x^5}. \]
Step 2: Rearrange the equation:
\[ \sin y \frac{dy}{dx} - \cos y \cdot \frac{7}{x} = -\frac{e^x}{x^5}. \]
Step 3: Let \( \cos y = t \), then:
\[ \sin y \frac{dy}{dx} = \frac{dt}{dx}. \]
Step 4: Substitute and simplify:
\[ \frac{dt}{dx} + \frac{7t}{x} = -\frac{e^x}{x^5}. \]
The integrating factor (I.F.) is:
\[ I.F. = x^7. \]
Step 5: Multiply by the integrating factor:
\[ x^7 \cdot \left( \frac{dt}{dx} + \frac{7t}{x} \right) = x^7 \cdot \left( -\frac{e^x}{x^5} \right). \]
Which simplifies to:
\[ \cos y \cdot x^7 = \int x^2 e^x \, dx. \]
Step 6: Solve the integral:
\[ \cos y \cdot x^7 = x^2 e^x - 2 \int x e^x \, dx. \]
Further simplifying:
\[ \cos x^7 = x^2 e^x - 2 x e^x + 2 e^x + c. \]
Step 7: Final solution:
Substitute \( x = 1 \), \( y = \frac{\pi}{2} \), and \( c = -e \): \[ \cos y = \frac{2e^2 - e}{128}. \]
Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)