Question ID: #787
Let $y=y(x)$ be the solution of the differential equation $x^4 dy + (4x^3y + 2\sin x)dx = 0$, $x>0$ and $y\left(\frac{\pi}{2}\right)=0$. Then $\pi^4 y\left(\frac{\pi}{3}\right)$ is equal to:
- (1) 81
- (2) 92
- (3) 64
- (4) 72
Solution:
Rearrange the given differential equation:
$$ x^4 dy + (4x^3y)dx = -2\sin x dx $$
Notice that the Left Hand Side is the exact differential of $x^4y$:
$$ d(x^4y) = -2\sin x dx $$
Integrate both sides:
$$ \int d(x^4y) = \int -2\sin x dx $$
$$ x^4y = 2\cos x + C $$
Apply the initial condition $y\left(\frac{\pi}{2}\right)=0$:
$$ \left(\frac{\pi}{2}\right)^4 \cdot 0 = 2\cos\left(\frac{\pi}{2}\right) + C $$
$$ 0 = 0 + C \Rightarrow C = 0 $$
The particular solution is:
$$ x^4y = 2\cos x $$
We need to find the value of $\pi^4 y\left(\frac{\pi}{3}\right)$.
First, substitute $x = \frac{\pi}{3}$ into the solution:
$$ \left(\frac{\pi}{3}\right)^4 y\left(\frac{\pi}{3}\right) = 2\cos\left(\frac{\pi}{3}\right) $$
$$ \frac{\pi^4}{81} y\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} $$
$$ \frac{\pi^4}{81} y\left(\frac{\pi}{3}\right) = 1 $$
Multiply both sides by 81:
$$ \pi^4 y\left(\frac{\pi}{3}\right) = 81 $$
Ans. (1)
Was this solution helpful?
YesNo