$x\frac{dy}{dx} - \sin 2y = x^3(2-x^3)\cos^2y$, $x \neq 0$.
If $y(2) = 0$, then $\tan(y(1))$ is equal to
Differential Equation Transformation
The given differential equation is:
$x\frac{dy}{dx} - \sin 2y = x^3(2-x^3)\cos^2y$
Since $x \neq 0$, we can divide by $\cos^2y$ (assuming $\cos y \neq 0$).
$x\frac{1}{\cos^2y}\frac{dy}{dx} - \frac{\sin 2y}{\cos^2y} = x^3(2-x^3)$
Using $\sec^2y = \frac{1}{\cos^2y}$ and $\sin 2y = 2\sin y \cos y$, we get:
$x\sec^2y\frac{dy}{dx} - \frac{2\sin y \cos y}{\cos^2y} = x^3(2-x^3)$
$x\sec^2y\frac{dy}{dx} - 2\frac{\sin y}{\cos y} = x^3(2-x^3)$
$x\sec^2y\frac{dy}{dx} - 2\tan y = x^3(2-x^3)$
Substitution for Linearization
Let $t = \tan y$. Then, differentiating with respect to $x$ gives:
$\frac{dt}{dx} = \sec^2y \frac{dy}{dx}$
Substituting $t$ and $\frac{dt}{dx}$ into the equation:
$x\frac{dt}{dx} - 2t = x^3(2-x^3)$
This is a linear first-order ordinary differential equation in $t$. Rearranging it:
$\frac{dt}{dx} - \frac{2}{x}t = x^2(2-x^3)$
Solving the Linear ODE
The integrating factor (IF) is:
$IF = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = e^{\ln(x^{-2})} = x^{-2}$
Multiply the linear ODE by the integrating factor:
$x^{-2}\frac{dt}{dx} - 2x^{-3}t = x^{-2} \cdot x^2(2-x^3)$
The left side is the derivative of the product $(t \cdot IF)$:
$\frac{d}{dx}(t x^{-2}) = 2 - x^3$
Integrate both sides with respect to $x$:
$\int \frac{d}{dx}(t x^{-2}) dx = \int (2 - x^3) dx$
$t x^{-2} = 2x - \frac{x^4}{4} + C$
Where $C$ is the constant of integration.
Solve for $t$:
$t = x^2 \left( 2x - \frac{x^4}{4} + C \right) = 2x^3 - \frac{x^6}{4} + C x^2$
Applying the Initial Condition
Substitute back $t = \tan y$:
$\tan y = 2x^3 - \frac{x^6}{4} + C x^2$
We are given the condition $y(2) = 0$. Substitute $x=2$ and $y=0$:
$\tan(0) = 2(2^3) - \frac{2^6}{4} + C (2^2)$
$0 = 2(8) - \frac{64}{4} + C(4)$
$0 = 16 - 16 + 4C$
$0 = 4C \implies C = 0$
Finding the Final Value
The specific solution is:
$\tan y = 2x^3 - \frac{x^6}{4}$
We need to find $\tan(y(1))$. Substitute $x=1$ into the specific solution:
$\tan(y(1)) = 2(1^3) - \frac{1^6}{4}$
$\tan(y(1)) = 2(1) - \frac{1}{4}$
$\tan(y(1)) = 2 - \frac{1}{4}$
$\tan(y(1)) = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}$
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Important Questions from Calculus
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