Question

If \[ \frac{dy}{dx} = 8y^2x^3 \] and \(y(2)=1\), then \(\dfrac{1}{y(0)}\) (in integer) is ......................

Hint:

Always use separation of variables in such first-order ODEs. Don’t forget to apply the initial condition after integration to determine the constant of integration.

Answer 1

Step 1: Separate the differential equation
\[ \frac{dy}{dx} = 8y^2x^3 \] This is separable: \[ \frac{dy}{y^2} = 8x^3 dx \] Step 2: Integrate both sides
\[ \int \frac{dy}{y^2} = \int 8x^3 dx \] \[ - \frac{1}{y} = 2x^4 + C \] Step 3: Apply initial condition $y(2) = 1$
\[ - \frac{1}{1} = 2(2^4) + C \] \[ -1 = 32 + C ⇒ C = -33 \] Step 4: General solution
\[ - \frac{1}{y} = 2x^4 - 33 \] \[ \frac{1}{y} = 33 - 2x^4 \] Step 5: Find $\dfrac{1{y(0)}$}
At $x=0$: \[ \frac{1}{y(0)} = 33 - 2(0)^4 = 33 \] Final Answer: \[ \frac{1}{y(0)} = 33 \]