If y is the solution of the differential equation , y(0)=1, the value of y(-1) is

Correct Answer: 0

Step-by-step Explanation:

By analyzing the provided image, we can identify the first-order ordinary differential equation written on the screen as:

$y^3\frac{dy}{dx}+x^3=0$

We are given the initial condition $y\left(0\right)=1$, and we need to find the value of $y(-1)$.

Step 1: Separate the variables
We rearrange the differential equation to group all terms containing $y$ on one side and all terms containing $x$ on the other side:

$y^3\frac{dy}{dx}=-x^3$

Multiplying both sides by $dx$, we obtain the separated form:

$y^{3}dy=-x^{3}dx$

Step 2: Integrate both sides
Now, we integrate both sides of the equation to find the general solution:

$\int y^{3}dy=-\int x^{3}dx$

Using the power rule for integration, $\int u^{n}du=\frac{u^{n+1}}{n+1}$, we get:

$\frac{y^4}{4}=-\frac{x^4}{4}+C$

where $C$ is the constant of integration. Multiplying the entire equation by 4 gives:

$y^4=-x^4+K$

where $K=4C$ is an arbitrary constant. Rearranging terms, we have:

$y^4+x^4=K$

Step 3: Solve for the constant K using the initial condition
We are given that $y\left(0\right)=1$, meaning $y=1$ when $x=0$. Substituting these values into our equation:

${\left(1\right)}^4+{\left(0\right)}^4=K$

$K=1$

Thus, the particular solution of the differential equation is:

$y^4+x^4=1$

Step 4: Find the value of y(-1)
To determine $y(-1)$, we substitute $x=-1$ into the particular solution:

$y^4+{(-1)}^4=1$

Since ${(-1)}^4=1$, the equation simplifies to:

$y^4+1=1$

Subtracting 1 from both sides:

$y^4=0$

Taking the fourth root of both sides gives:

$y=0$

Therefore, the value of $y(-1)$ is 0.