If the solution curve of the differential equation \((2x - 10y^3)dy + y dx = 0\), passes through the points (0, 1) and (2, \(\beta\)), then \(\beta\) is a root of the equation :
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When a differential equation is not easily separable or homogeneous, check if it can be rearranged into a linear form, either \(dy/dx + P(x)y = Q(x)\) or \(dx/dy + P(y)x = Q(y)\). Identifying the correct form is the key to solving it using the integrating factor method.
Step 1: Rearrange the differential equation.
\[ (2x - 10y^3)dy + y dx = 0 \]
\[ y \frac{dx}{dy} + 2x - 10y^3 = 0 \]
\[ \frac{dx}{dy} + \frac{2}{y}x = 10y^2 \]
This is a linear differential equation of the form \(\frac{dx}{dy} + P(y)x = Q(y)\), where \(P(y) = 2/y\) and \(Q(y) = 10y^2\).
Step 2: Find the integrating factor (I.F.).
\[ I.F. = e^{\int P(y) dy} = e^{\int \frac{2}{y} dy} = e^{2\ln y} = e^{\ln y^2} = y^2 \]
Step 3: Find the general solution.
The solution is given by \(x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C\).
\[ x \cdot y^2 = \int (10y^2) \cdot y^2 dy + C \]
\[ xy^2 = \int 10y^4 dy + C \]
\[ xy^2 = 10 \frac{y^5}{5} + C = 2y^5 + C \]
Step 4: Use the initial condition to find C.
The curve passes through (0, 1). So, x=0, y=1.
\[ (0)(1)^2 = 2(1)^5 + C \implies 0 = 2 + C \implies C = -2 \]
The particular solution is \(xy^2 = 2y^5 - 2\).
Step 5: Use the second point to find the equation for \(\beta\).
The curve also passes through (2, \(\beta\)). So, x=2, y=\(\beta\).
\[ (2)(\beta)^2 = 2(\beta)^5 - 2 \]
Divide by 2:
\[ \beta^2 = \beta^5 - 1 \]
Rearranging the terms gives the equation that \(\beta\) must satisfy:
\[ \beta^5 - \beta^2 - 1 = 0 \]
This means \(\beta\) is a root of the equation \(y^5 - y^2 - 1 = 0\).
