Ifdy/dx = x - y2 and y(0) = 1, then y(0.1) correct upto two decimal places (approx.) is:a)0.85b)0.84c)0.91d)1.01Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Concept:
Euler's Method to generate a numerical solution to an initial value problem of the form:
y' = f (x, y), y(x₀) = y₀
y_{n + 1} = y_n + h f(x_n, y_n)
Calculation:
We have,
x₀ = 0, y₀ = 1, h = 0.1
x₁ = x₀ + h
For n = 0
⇒ y₁ = y₀ + hf(x_n, y_n)
⇒ y₁ = 1 - 0.1 × (0 - 1)²
⇒ y₁ = 1 - 0.1
⇒ y₁ = 0. 9
For n = 1
⇒ y₂ = y₁ + hf(x₁ - y₁²)
⇒ y₂ = 0.9 + 0.1[0.1 - (0.9)²]
⇒ y₂ = 0.9 + 0.1[0.1 - 0. 81]
⇒ y₂ = 0.9 - 0.1 × 0.71
⇒ y₂ = 0.9 - 0.071
⇒ y₂ = 0. 829
⇒ y₂ = 0.83
∴ The approximately. two decimal place value is 0.83

Answer

What is the order of time required to fit a spline curve to n points?

a.

O(n)

Answer

Given:
$\frac{dy}{dx} = x - y^2$
$y(0) = 1$

To Find:
$y(0.1)$ correct up to two decimal places.

Explanation:
To solve this differential equation, we can use the method of separation of variables.

Separating the variables, we get:
$\frac{dy}{1-y^2} = x \, dx$

Integration:
Integrating both sides with respect to their respective variables, we get:
$\int{\frac{dy}{1-y^2}} = \int{x \, dx}$

LHS Integration:
To integrate the left-hand side, we can use partial fraction decomposition. The general form of the partial fraction decomposition is:
$\frac{A}{y-1} + \frac{B}{y+1}$

Multiplying through by the common denominator $(y-1)(y+1)$, we get:
$1 = A(y+1) + B(y-1)$

Expanding and equating the coefficients of like terms, we get:
$1 = (A + B)y + (A - B)$

Comparing the coefficients of 'y', we get:
$A + B = 0 \implies A = -B$

Comparing the constants, we get:
$A - B = 1$

Solving these equations, we find:
$A = \frac{1}{2}$ and $B = -\frac{1}{2}$

Substituting these values back into the partial fraction decomposition, we get:
$\frac{\frac{1}{2}}{y-1} - \frac{\frac{1}{2}}{y+1}$

Integrating each term separately, we get:
$\frac{1}{2}\ln|y-1| - \frac{1}{2}\ln|y+1|$

RHS Integration:
Integrating the right-hand side, we get:
$\int{x \, dx} = \frac{x^2}{2} + C_1$

Combining the Integrals:
Substituting the integrals back into the original equation, we get:
$\frac{1}{2}\ln|y-1| - \frac{1}{2}\ln|y+1| = \frac{x^2}{2} + C_1$

Simplifying the equation, we get:
$\ln\left|\frac{y-1}{y+1}\right| = x^2 + C_1$

Applying Initial Condition:
Now, we can apply the initial condition $y(0) = 1$.

Plugging in the values, we get:
$\ln\left|\frac{1-1}{1+1}\right| = 0^2 + C_1$

Simplifying, we get:
$\ln\left|\frac{0}{2}\right| = C_1$

Which further simplifies to:
$\ln(0) = C_1$

The natural logarithm of 0 is undefined. Therefore, $C_1$ is undefined.

Solving for y:
Now, we can solve the equation for y.

Taking the exponent of both sides, we get:

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