Questions: Consider the following Bernoulli equation, [y'+fracyx-sqrty=0] Find an explicit form of the general solution, and also find and graph the particular solution for the case (y(1)=0).

Solution Steps

The given differential equation is a Bernoulli equation of the form:

\[ y^{\prime} + P(x)y = Q(x)y^n \]

where \( n = \frac{1}{2} \), \( P(x) = \frac{1}{x} \), and \( Q(x) = -1 \).

To solve the Bernoulli equation, we perform the substitution \( v = y^{1-n} = y^{\frac{1}{2}} \). Then, \( y = v^2 \) and \( y' = 2v v' \).

Substituting these into the original equation gives:

\[ 2v v' + \frac{v^2}{x} - v = 0 \]

Divide the entire equation by \( 2v \):

\[ v' + \frac{v}{2x} = \frac{1}{2} \]

This is a linear first-order differential equation. The integrating factor is:

\[ \mu(x) = e^{\int \frac{1}{2x} \, dx} = e^{\frac{1}{2} \ln x} = x^{1/2} \]

Multiply through by the integrating factor:

\[ x^{1/2} v' + \frac{v x^{1/2}}{2x} = \frac{x^{1/2}}{2} \]

\[ \frac{d}{dx}(x^{1/2} v) = \frac{x^{1/2}}{2} \]

Integrate both sides:

\[ x^{1/2} v = \int \frac{x^{1/2}}{2} \, dx = \frac{1}{3} x^{3/2} + C \]

Solve for \( v \):

\[ v = \frac{1}{3} x + C x^{-1/2} \]

Since \( v = y^{1/2} \), we have:

\[ y = \left(\frac{1}{3} x + C x^{-1/2}\right)^2 \]

Given \( y(1) = 0 \):

\[ 0 = \left(\frac{1}{3} \cdot 1 + C \cdot 1^{-1/2}\right)^2 \]

\[ 0 = \left(\frac{1}{3} + C\right)^2 \]

Thus, \( C = -\frac{1}{3} \).

The particular solution is:

\[ y = \left(\frac{1}{3} x - \frac{1}{3} x^{-1/2}\right)^2 \]

Final Answer