Solved on Nov 01, 2023

Find the solution of the initial value problem $g'(x)=8x(x^7-1/8); g(1)=1$ . The solution is $g(x)=\square$ .

STEP 1

Assumptions1. The function $g(x)$ is differentiable. . The derivative of $g(x)$ is given by $g'(x)=8x(x^7-\frac{1}{8})$ .
3. The initial condition is $g(1)=1$ .

STEP 2

First, we need to solve the differential equation $g'(x)=8x(x^7-\frac{1}{8})$ . This is a first-order ordinary differential equation, and we can solve it by integrating both sides with respect to $x$ .
$\int g'(x) \, dx = \int8x(x^7-\frac{1}{8}) \, dx$

STEP 3

The integral of $g'(x)$ with respect to $x$ is simply $g(x)$ , so we have $g(x) = \int8x(x^7-\frac{1}{8}) \, dx$

STEP 4

We can simplify the integral on the right-hand side by distributing the $8x$ inside the parentheses.
$g(x) = \int8x^8 - x \, dx$

STEP 5

Now we can integrate term by term.
$g(x) = \int8x^8 \, dx - \int x \, dx$

STEP 6

The integral of $8x^8$ with respect to $x$ is $x^9$ and the integral of $x$ with respect to $x$ is $\frac{1}{2}x^2$ . So we have $g(x) = x^9 - \frac{1}{2}x^2 + C$ where $C$ is the constant of integration.

STEP 7

Now we use the initial condition $g(1)=1$ to find the value of $C$ . Substituting $x=1$ into the equation gives us $1 =1^9 - \frac{1}{2}1^2 + C$

STEP 8

olving for $C$ gives us $C =1 -1 + \frac{1}{2} = \frac{1}{2}$

STEP 9

Substituting $C=\frac{}{2}$ back into the equation for $g(x)$ gives us the solution to the initial value problem $g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}$ So, the solution is $g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}$ .

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Find the solution of the initial value problem $g'(x)=8x(x^7-1/8); g(1)=1$ . The solution is $g(x)=\square$ .

STEP 1

Assumptions1. The function $g(x)$ is differentiable. . The derivative of $g(x)$ is given by $g'(x)=8x(x^7-\frac{1}{8})$ .
3. The initial condition is $g(1)=1$ .

STEP 2

First, we need to solve the differential equation $g'(x)=8x(x^7-\frac{1}{8})$ . This is a first-order ordinary differential equation, and we can solve it by integrating both sides with respect to $x$ .
$\int g'(x) \, dx = \int8x(x^7-\frac{1}{8}) \, dx$

STEP 3

The integral of $g'(x)$ with respect to $x$ is simply $g(x)$ , so we have $g(x) = \int8x(x^7-\frac{1}{8}) \, dx$

STEP 4

We can simplify the integral on the right-hand side by distributing the $8x$ inside the parentheses.
$g(x) = \int8x^8 - x \, dx$

STEP 5

Now we can integrate term by term.
$g(x) = \int8x^8 \, dx - \int x \, dx$

STEP 6

The integral of $8x^8$ with respect to $x$ is $x^9$ and the integral of $x$ with respect to $x$ is $\frac{1}{2}x^2$ . So we have $g(x) = x^9 - \frac{1}{2}x^2 + C$ where $C$ is the constant of integration.

STEP 7

Now we use the initial condition $g(1)=1$ to find the value of $C$ . Substituting $x=1$ into the equation gives us $1 =1^9 - \frac{1}{2}1^2 + C$

STEP 8

olving for $C$ gives us $C =1 -1 + \frac{1}{2} = \frac{1}{2}$

STEP 9

Substituting $C=\frac{}{2}$ back into the equation for $g(x)$ gives us the solution to the initial value problem $g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}$ So, the solution is $g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}$ .

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Find the solution of the initial value problem g′(x)=8x(x7−1/8);g(1)=1g'(x)=8x(x^7-1/8); g(1)=1g′(x)=8x(x7−1/8);g(1)=1. The solution is g(x)=□g(x)=\squareg(x)=□.

STEP 1

Assumptions1. The function g(x)g(x)g(x) is differentiable. . The derivative of g(x)g(x)g(x) is given by g′(x)=8x(x7−18)g'(x)=8x(x^7-\frac{1}{8})g′(x)=8x(x7−81​).3. The initial condition is g(1)=1g(1)=1g(1)=1.

STEP 2

STEP 3

The integral of g′(x)g'(x)g′(x) with respect to xxx is simply g(x)g(x)g(x), so we haveg(x)=∫8x(x7−18) dxg(x) = \int8x(x^7-\frac{1}{8}) \, dxg(x)=∫8x(x7−81​)dx

STEP 4

We can simplify the integral on the right-hand side by distributing the 8x8x8x inside the parentheses.g(x)=∫8x8−x dxg(x) = \int8x^8 - x \, dxg(x)=∫8x8−xdx

STEP 5

Now we can integrate term by term.g(x)=∫8x8 dx−∫x dxg(x) = \int8x^8 \, dx - \int x \, dxg(x)=∫8x8dx−∫xdx

STEP 6

The integral of 8x88x^88x8 with respect to xxx is x9x^9x9 and the integral of xxx with respect to xxx is 12x2\frac{1}{2}x^221​x2. So we haveg(x)=x9−12x2+Cg(x) = x^9 - \frac{1}{2}x^2 + Cg(x)=x9−21​x2+Cwhere CCC is the constant of integration.

STEP 7

Now we use the initial condition g(1)=1g(1)=1g(1)=1 to find the value of CCC. Substituting x=1x=1x=1 into the equation gives us1=19−1212+C1 =1^9 - \frac{1}{2}1^2 + C1=19−21​12+C

STEP 8

olving for CCC gives usC=1−1+12=12C =1 -1 + \frac{1}{2} = \frac{1}{2}C=1−1+21​=21​

STEP 9

Substituting C=2C=\frac{}{2}C=2​ back into the equation for g(x)g(x)g(x) gives us the solution to the initial value problemg(x)=x9−2x2+2g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}g(x)=x9−2​x2+2​So, the solution is g(x)=x9−2x2+2g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}g(x)=x9−2​x2+2​.

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Find the solution of the initial value problem $g'(x)=8x(x^7-1/8); g(1)=1$ . The solution is $g(x)=\square$ .

Assumptions1. The function $g(x)$ is differentiable. . The derivative of $g(x)$ is given by $g'(x)=8x(x^7-\frac{1}{8})$ .
3. The initial condition is $g(1)=1$ .

The integral of $g'(x)$ with respect to $x$ is simply $g(x)$ , so we have $g(x) = \int8x(x^7-\frac{1}{8}) \, dx$

We can simplify the integral on the right-hand side by distributing the $8x$ inside the parentheses.
$g(x) = \int8x^8 - x \, dx$

Now we can integrate term by term.
$g(x) = \int8x^8 \, dx - \int x \, dx$

The integral of $8x^8$ with respect to $x$ is $x^9$ and the integral of $x$ with respect to $x$ is $\frac{1}{2}x^2$ . So we have $g(x) = x^9 - \frac{1}{2}x^2 + C$ where $C$ is the constant of integration.

Now we use the initial condition $g(1)=1$ to find the value of $C$ . Substituting $x=1$ into the equation gives us $1 =1^9 - \frac{1}{2}1^2 + C$

olving for $C$ gives us $C =1 -1 + \frac{1}{2} = \frac{1}{2}$

Substituting $C=\frac{}{2}$ back into the equation for $g(x)$ gives us the solution to the initial value problem $g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}$ So, the solution is $g(x) = x^9 - \frac{}{2}x^2 + \frac{}{2}$ .