3. determine the equation of the solution curve to dy/dx = 2x(1 - 3y) that satisfies y(0)=1.

Explanation:

Step1: Separate variables

Separate the variables in the differential equation $\frac{dy}{dx}=2x(1 - 3y)$. We get $\frac{dy}{1 - 3y}=2x dx$.

Step2: Integrate both sides

Integrate $\int\frac{dy}{1 - 3y}=\int2x dx$. For $\int\frac{dy}{1 - 3y}$, let $u = 1-3y$, then $du=-3dy$ and $\int\frac{dy}{1 - 3y}=-\frac{1}{3}\int\frac{du}{u}=-\frac{1}{3}\ln|u|=-\frac{1}{3}\ln|1 - 3y|$. For $\int2x dx=x^{2}+C$. So, $-\frac{1}{3}\ln|1 - 3y|=x^{2}+C$.

Step3: Use the initial - condition

Given $y(0) = 1$, substitute $x = 0$ and $y = 1$ into $-\frac{1}{3}\ln|1 - 3y|=x^{2}+C$. $-\frac{1}{3}\ln|1-3\times1|=0^{2}+C$, $-\frac{1}{3}\ln|-2|=C$, $C =-\frac{1}{3}\ln2$.

Step4: Solve for y

We have $-\frac{1}{3}\ln|1 - 3y|=x^{2}-\frac{1}{3}\ln2$. Multiply through by - 3: $\ln|1 - 3y|=-3x^{2}+\ln2$. Exponentiate both sides: $|1 - 3y| = e^{-3x^{2}+\ln2}=2e^{-3x^{2}}$. Since when $x = 0,y = 1$, $1-3y<0$, so $-(1 - 3y)=2e^{-3x^{2}}$. $3y=1 + 2e^{-3x^{2}}$, $y=\frac{1 + 2e^{-3x^{2}}}{3}$.

Answer:

$y=\frac{1 + 2e^{-3x^{2}}}{3}$