Solved on Mar 02, 2024

Use Euler's method with step size 0.1 to find $y(0.2)$ , where $y(x)$ satisfies $y'=2x+3y^2, y(0)=0$ .

STEP 1

Assumptions
1. The differential equation is $y' = 2x + 3y^2$ .
2. The initial condition is $y(0) = 0$ .
3. We want to approximate $y(0.2)$ using Euler's method.
4. The step size for Euler's method is $h = 0.1$ .
5. Euler's method uses the formula $y_{n+1} = y_n + h \cdot f(x_n, y_n)$ , where $f(x, y) = y'$ .

STEP 2

We start by using the initial condition to establish our first point $(x_0, y_0)$ .
$x_0 = 0, \quad y_0 = 0$

STEP 3

Apply Euler's method to find the next approximation $y_1$ at $x_1 = x_0 + h$ .
$y_1 = y_0 + h \cdot f(x_0, y_0)$

STEP 4

Calculate $f(x_0, y_0)$ using the given differential equation.
$f(x_0, y_0) = 2x_0 + 3y_0^2$

STEP 5

Substitute $x_0 = 0$ and $y_0 = 0$ into the function $f(x_0, y_0)$ .
$f(0, 0) = 2 \cdot 0 + 3 \cdot 0^2 = 0$

STEP 6

Now we can calculate $y_1$ .
$y_1 = y_0 + h \cdot f(0, 0)$

STEP 7

Substitute $y_0 = 0$ and $h = 0.1$ into the equation for $y_1$ .
$y_1 = 0 + 0.1 \cdot 0 = 0$

STEP 8

Next, we need to find the approximation $y_2$ at $x_2 = x_1 + h$ .
$y_2 = y_1 + h \cdot f(x_1, y_1)$

STEP 9

Since $x_1 = x_0 + h$ , calculate $x_1$ .
$x_1 = 0 + 0.1 = 0.1$

STEP 10

Calculate $f(x_1, y_1)$ using the given differential equation.
$f(x_1, y_1) = 2x_1 + 3y_1^2$

STEP 11

Substitute $x_1 = 0.1$ and $y_1 = 0$ into the function $f(x_1, y_1)$ .
$f(0.1, 0) = 2 \cdot 0.1 + 3 \cdot 0^2 = 0.2$

STEP 12

Now we can calculate $y_2$ .
$y_2 = y_1 + h \cdot f(0.1, 0)$

STEP 13

Substitute $y_1 = 0$ and $h = 0.1$ into the equation for $y_2$ .
$y_2 = 0 + 0.1 \cdot 0.2 = 0.02$

STEP 14

Since we want to approximate $y(0.2)$ , we have reached our desired $x$ value at $x_2 = 0.2$ . Therefore, our approximation for $y(0.2)$ is $y_2$ .
$y(0.2) \approx y_2 = 0.02$

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Use Euler's method with step size 0.1 to find $y(0.2)$ , where $y(x)$ satisfies $y'=2x+3y^2, y(0)=0$ .

STEP 1

STEP 2

We start by using the initial condition to establish our first point $(x_0, y_0)$ .
$x_0 = 0, \quad y_0 = 0$

STEP 3

Apply Euler's method to find the next approximation $y_1$ at $x_1 = x_0 + h$ .
$y_1 = y_0 + h \cdot f(x_0, y_0)$

STEP 4

Calculate $f(x_0, y_0)$ using the given differential equation.
$f(x_0, y_0) = 2x_0 + 3y_0^2$

STEP 5

Substitute $x_0 = 0$ and $y_0 = 0$ into the function $f(x_0, y_0)$ .
$f(0, 0) = 2 \cdot 0 + 3 \cdot 0^2 = 0$

STEP 6

Now we can calculate $y_1$ .
$y_1 = y_0 + h \cdot f(0, 0)$

STEP 7

Substitute $y_0 = 0$ and $h = 0.1$ into the equation for $y_1$ .
$y_1 = 0 + 0.1 \cdot 0 = 0$

STEP 8

Next, we need to find the approximation $y_2$ at $x_2 = x_1 + h$ .
$y_2 = y_1 + h \cdot f(x_1, y_1)$

STEP 9

Since $x_1 = x_0 + h$ , calculate $x_1$ .
$x_1 = 0 + 0.1 = 0.1$

STEP 10

Calculate $f(x_1, y_1)$ using the given differential equation.
$f(x_1, y_1) = 2x_1 + 3y_1^2$

STEP 11

Substitute $x_1 = 0.1$ and $y_1 = 0$ into the function $f(x_1, y_1)$ .
$f(0.1, 0) = 2 \cdot 0.1 + 3 \cdot 0^2 = 0.2$

STEP 12

Now we can calculate $y_2$ .
$y_2 = y_1 + h \cdot f(0.1, 0)$

STEP 13

Substitute $y_1 = 0$ and $h = 0.1$ into the equation for $y_2$ .
$y_2 = 0 + 0.1 \cdot 0.2 = 0.02$

STEP 14

Since we want to approximate $y(0.2)$ , we have reached our desired $x$ value at $x_2 = 0.2$ . Therefore, our approximation for $y(0.2)$ is $y_2$ .
$y(0.2) \approx y_2 = 0.02$

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Use Euler's method with step size 0.1 to find y(0.2)y(0.2)y(0.2), where y(x)y(x)y(x) satisfies y′=2x+3y2,y(0)=0y'=2x+3y^2, y(0)=0y′=2x+3y2,y(0)=0.

STEP 1

STEP 2

We start by using the initial condition to establish our first point (x0,y0)(x_0, y_0)(x0​,y0​).x0=0,y0=0x_0 = 0, \quad y_0 = 0x0​=0,y0​=0

STEP 3

Apply Euler's method to find the next approximation y1y_1y1​ at x1=x0+hx_1 = x_0 + hx1​=x0​+h.y1=y0+h⋅f(x0,y0)y_1 = y_0 + h \cdot f(x_0, y_0)y1​=y0​+h⋅f(x0​,y0​)

STEP 4

Calculate f(x0,y0)f(x_0, y_0)f(x0​,y0​) using the given differential equation.f(x0,y0)=2x0+3y02f(x_0, y_0) = 2x_0 + 3y_0^2f(x0​,y0​)=2x0​+3y02​

STEP 5

Substitute x0=0x_0 = 0x0​=0 and y0=0y_0 = 0y0​=0 into the function f(x0,y0)f(x_0, y_0)f(x0​,y0​).f(0,0)=2⋅0+3⋅02=0f(0, 0) = 2 \cdot 0 + 3 \cdot 0^2 = 0f(0,0)=2⋅0+3⋅02=0

STEP 6

Now we can calculate y1y_1y1​.y1=y0+h⋅f(0,0)y_1 = y_0 + h \cdot f(0, 0)y1​=y0​+h⋅f(0,0)

STEP 7

Substitute y0=0y_0 = 0y0​=0 and h=0.1h = 0.1h=0.1 into the equation for y1y_1y1​.y1=0+0.1⋅0=0y_1 = 0 + 0.1 \cdot 0 = 0y1​=0+0.1⋅0=0

STEP 8

Next, we need to find the approximation y2y_2y2​ at x2=x1+hx_2 = x_1 + hx2​=x1​+h.y2=y1+h⋅f(x1,y1)y_2 = y_1 + h \cdot f(x_1, y_1)y2​=y1​+h⋅f(x1​,y1​)

STEP 9

Since x1=x0+hx_1 = x_0 + hx1​=x0​+h, calculate x1x_1x1​.x1=0+0.1=0.1x_1 = 0 + 0.1 = 0.1x1​=0+0.1=0.1

STEP 10

Calculate f(x1,y1)f(x_1, y_1)f(x1​,y1​) using the given differential equation.f(x1,y1)=2x1+3y12f(x_1, y_1) = 2x_1 + 3y_1^2f(x1​,y1​)=2x1​+3y12​

STEP 11

Substitute x1=0.1x_1 = 0.1x1​=0.1 and y1=0y_1 = 0y1​=0 into the function f(x1,y1)f(x_1, y_1)f(x1​,y1​).f(0.1,0)=2⋅0.1+3⋅02=0.2f(0.1, 0) = 2 \cdot 0.1 + 3 \cdot 0^2 = 0.2f(0.1,0)=2⋅0.1+3⋅02=0.2

STEP 12

Now we can calculate y2y_2y2​.y2=y1+h⋅f(0.1,0)y_2 = y_1 + h \cdot f(0.1, 0)y2​=y1​+h⋅f(0.1,0)

STEP 13

Substitute y1=0y_1 = 0y1​=0 and h=0.1h = 0.1h=0.1 into the equation for y2y_2y2​.y2=0+0.1⋅0.2=0.02y_2 = 0 + 0.1 \cdot 0.2 = 0.02y2​=0+0.1⋅0.2=0.02

STEP 14

Since we want to approximate y(0.2)y(0.2)y(0.2), we have reached our desired xxx value at x2=0.2x_2 = 0.2x2​=0.2. Therefore, our approximation for y(0.2)y(0.2)y(0.2) is y2y_2y2​.y(0.2)≈y2=0.02y(0.2) \approx y_2 = 0.02y(0.2)≈y2​=0.02

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Use Euler's method with step size 0.1 to find $y(0.2)$ , where $y(x)$ satisfies $y'=2x+3y^2, y(0)=0$ .

We start by using the initial condition to establish our first point $(x_0, y_0)$ .
$x_0 = 0, \quad y_0 = 0$

Apply Euler's method to find the next approximation $y_1$ at $x_1 = x_0 + h$ .
$y_1 = y_0 + h \cdot f(x_0, y_0)$

Calculate $f(x_0, y_0)$ using the given differential equation.
$f(x_0, y_0) = 2x_0 + 3y_0^2$

Substitute $x_0 = 0$ and $y_0 = 0$ into the function $f(x_0, y_0)$ .
$f(0, 0) = 2 \cdot 0 + 3 \cdot 0^2 = 0$

Now we can calculate $y_1$ .
$y_1 = y_0 + h \cdot f(0, 0)$

Substitute $y_0 = 0$ and $h = 0.1$ into the equation for $y_1$ .
$y_1 = 0 + 0.1 \cdot 0 = 0$

Next, we need to find the approximation $y_2$ at $x_2 = x_1 + h$ .
$y_2 = y_1 + h \cdot f(x_1, y_1)$

Since $x_1 = x_0 + h$ , calculate $x_1$ .
$x_1 = 0 + 0.1 = 0.1$

Calculate $f(x_1, y_1)$ using the given differential equation.
$f(x_1, y_1) = 2x_1 + 3y_1^2$

Substitute $x_1 = 0.1$ and $y_1 = 0$ into the function $f(x_1, y_1)$ .
$f(0.1, 0) = 2 \cdot 0.1 + 3 \cdot 0^2 = 0.2$

Now we can calculate $y_2$ .
$y_2 = y_1 + h \cdot f(0.1, 0)$

Substitute $y_1 = 0$ and $h = 0.1$ into the equation for $y_2$ .
$y_2 = 0 + 0.1 \cdot 0.2 = 0.02$

Since we want to approximate $y(0.2)$ , we have reached our desired $x$ value at $x_2 = 0.2$ . Therefore, our approximation for $y(0.2)$ is $y_2$ .
$y(0.2) \approx y_2 = 0.02$