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Step-by-Step Solution
Step 1: Rewrite the given differential equation
The given differential equation is:
$$(2x - 10y^3)\,dy + y\,dx = 0.$$
We aim to put this in the form:
$$\frac{dx}{dy} + P(y)\, x = Q(y).$$
First, isolate $dx$:
$$
y \, dx = - (2x - 10y^3)\, dy
\quad \Longrightarrow \quad
dx = - \frac{(2x - 10y^3)}{y}\, dy.
$$
This can be rearranged as:
$$
\frac{dx}{dy} + \frac{2}{y}\, x = 10y^2.
$$
Step 2: Determine the Integrating Factor (I.F.)
For a linear differential equation of the form
$$\frac{dx}{dy} + P(y)\,x = Q(y),$$
the integrating factor is given by
$$
\text{I.F.} = e^{\int P(y)\, dy}.
$$
Here, $P(y) = \frac{2}{y}$. Thus,
$$
\int P(y)\,dy = \int \frac{2}{y}\,dy = 2 \ln(y).
$$
Therefore,
$$
\text{I.F.} = e^{2\ln(y)} = y^2.
$$
Step 3: Multiply through by the Integrating Factor
We multiply the entire differential equation by $y^2$:
$$
y^2 \frac{dx}{dy} + \left(\frac{2}{y} \cdot y^2\right) x = 10y^2 \cdot y^2.
$$
This simplifies to:
$$
y^2 \frac{dx}{dy} + 2y\,x = 10y^4.
$$
Notice that the left-hand side can be written as the derivative of $(x\,y^2)$ with respect to $y$:
$$
\frac{d}{dy} \bigl(x\,y^2\bigr) = 10y^4.
$$
Step 4: Integrate both sides
Integrate with respect to $y$:
$$
x\,y^2 = \int 10y^4 \, dy.
$$
The right-hand side becomes:
$$
\int 10y^4 \, dy = 10 \cdot \frac{y^5}{5} = 2y^5.
$$
So the general solution is:
$$
x\,y^2 = 2y^5 + C,
$$
where $C$ is the constant of integration.
Step 5: Use the initial condition (0, 1)
The solution curve passes through $(0, 1)$. Substitute $x = 0$ and $y = 1$:
$$
0 \cdot (1)^2 = 2(1)^5 + C
\quad \Longrightarrow \quad
0 = 2 + C
\quad \Longrightarrow \quad
C = -2.
$$
Thus, the particular solution becomes:
$$
x\,y^2 = 2y^5 - 2.
$$
Step 6: Apply the second condition (2, β)
The curve also passes through $(2, \beta)$. Substitute $x = 2$ and $y = \beta$ into
$x\,y^2 = 2y^5 - 2$:
$$
2\,\beta^2 = 2\,\beta^5 - 2.
$$
Rearrange:
$$
2\,\beta^5 - 2\,\beta^2 - 2 = 0
\quad \Longrightarrow \quad
2\,\beta^5 = 2\,\beta^2 + 2
\quad \Longrightarrow \quad
\beta^5 = \beta^2 + 1
\quad \Longrightarrow \quad
\beta^5 - \beta^2 - 1 = 0.
$$
Step 7: Conclusion
Hence, $\beta$ satisfies the equation:
$$
y^5 - y^2 - 1 = 0.
$$
This matches the correct answer:
y5 − y2 − 1 = 0.
