For the equation, , if y(0) = 3/7,then the value of y(1) is

The correct answer is:
$$\frac{3}{7}{e}^{-\frac{7}{3}}$$

Step-by-Step Explanation:

1. Identify the Differential Equation and Given Conditions
From the provided question image, we are given the first-order homogeneous ordinary differential equation:
$$\frac{dy}{dx}+7x^{2}y=0$$
We are also given the initial condition:
$$y\left(0\right)=\frac{3}{7}$$
We need to find the value of $y\left(1\right)$.

2. Separate the Variables
This equation is separable. We can rearrange it to group all terms with $y$ on one side and all terms with $x$ on the other side:
$$\frac{dy}{dx}=-7x^{2}y$$
Dividing both sides by $y$ (assuming $y\ne 0$) and multiplying by $dx$, we obtain:
$$\frac{dy}{y}=-7x^{2}dx$$

3. Integrate Both Sides
Now, integrate both sides to solve for $y$:
$$\int \frac{1}{y}dy=\int -7x^{2}dx$$
Integrating, we get:
$$\ln \left|y\right|=-7\left(\frac{x^3}{3}\right)+C$$
where $C$ is the constant of integration.
Simplifying:
$$\ln \left|y\right|=-\frac{7x^3}{3}+C$$

4. Exponentiate to Find the General Solution
To solve for $y$, we exponentiate both sides:
$$\left|y\right|=e^{-\frac{7x^3}{3}+C}$$
Using exponent properties:
$$y\left(x\right)=A{e}^{-\frac{7x^3}{3}}$$
where $A=\pm e^C$ is a constant.

5. Apply the Initial Condition
We are given the initial condition $y\left(0\right)=\frac{3}{7}$. Substituting $x=0$ into our general solution:
$$y\left(0\right)=A{e}^0=A=\frac{3}{7}$$
Thus, our particular solution is:
$$y\left(x\right)=\frac{3}{7}{e}^{-\frac{7x^3}{3}}$$

6. Calculate the Value of $y\left(1\right)$
Substitute $x=1$ into the particular solution:
$$y\left(1\right)=\frac{3}{7}{e}^{-\frac{7{\left(1\right)}^3}{3}}$$
Simplifying the exponent:
$$y\left(1\right)=\frac{3}{7}{e}^{-\frac{7}{3}}$$
This matches the expression shown in the correct option image.