Math / Calculus

QuestionIn Exercises 47-60, solve the initial value problem.
47. $\frac{d y}{d x}=x^{3}, y(0)=4$

Studdy Solution

STEP 1

1. The differential equation given is separable and can be solved by integration.
2. The initial condition $y(0) = 4$ will be used to find the constant of integration.

STEP 2

1. Integrate the differential equation to find the general solution.
2. Apply the initial condition to determine the constant of integration.
3. Write the particular solution that satisfies the initial condition.

STEP 3

Integrate the differential equation $\frac{d y}{d x} = x^3$ with respect to $x$ :
$\int \frac{d y}{d x} \, dx = \int x^3 \, dx$
This simplifies to:
$y = \frac{x^4}{4} + C$
where $C$ is the constant of integration.

STEP 4

Apply the initial condition $y(0) = 4$ to find the constant $C$ :
Substitute $x = 0$ and $y = 4$ into the general solution:
$4 = \frac{0^4}{4} + C$ $4 = 0 + C$ $C = 4$

STEP 5

Substitute the value of $C$ back into the general solution to find the particular solution:
$y = \frac{x^4}{4} + 4$
This is the solution to the initial value problem.
The particular solution is:
$y = \frac{x^4}{4} + 4$

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QuestionIn Exercises 47-60, solve the initial value problem.
47. $\frac{d y}{d x}=x^{3}, y(0)=4$

Studdy Solution

STEP 1

1. The differential equation given is separable and can be solved by integration.
2. The initial condition $y(0) = 4$ will be used to find the constant of integration.

STEP 2

1. Integrate the differential equation to find the general solution.
2. Apply the initial condition to determine the constant of integration.
3. Write the particular solution that satisfies the initial condition.

STEP 3

Integrate the differential equation $\frac{d y}{d x} = x^3$ with respect to $x$ :
$\int \frac{d y}{d x} \, dx = \int x^3 \, dx$
This simplifies to:
$y = \frac{x^4}{4} + C$
where $C$ is the constant of integration.

STEP 4

Apply the initial condition $y(0) = 4$ to find the constant $C$ :
Substitute $x = 0$ and $y = 4$ into the general solution:
$4 = \frac{0^4}{4} + C$ $4 = 0 + C$ $C = 4$

STEP 5

Substitute the value of $C$ back into the general solution to find the particular solution:
$y = \frac{x^4}{4} + 4$
This is the solution to the initial value problem.
The particular solution is:
$y = \frac{x^4}{4} + 4$

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QuestionIn Exercises 47-60, solve the initial value problem.47. dydx=x3,y(0)=4\frac{d y}{d x}=x^{3}, y(0)=4dxdy​=x3,y(0)=4

Studdy Solution

STEP 1

1. The differential equation given is separable and can be solved by integration.2. The initial condition y(0)=4 y(0) = 4 y(0)=4 will be used to find the constant of integration.

STEP 2

1. Integrate the differential equation to find the general solution.2. Apply the initial condition to determine the constant of integration.3. Write the particular solution that satisfies the initial condition.

STEP 3

Integrate the differential equation dydx=x3\frac{d y}{d x} = x^3dxdy​=x3 with respect to xxx:∫dydx dx=∫x3 dx \int \frac{d y}{d x} \, dx = \int x^3 \, dx ∫dxdy​dx=∫x3dxThis simplifies to:y=x44+C y = \frac{x^4}{4} + C y=4x4​+Cwhere C C C is the constant of integration.

STEP 4

Apply the initial condition y(0)=4 y(0) = 4 y(0)=4 to find the constant C C C:Substitute x=0 x = 0 x=0 and y=4 y = 4 y=4 into the general solution:4=044+C 4 = \frac{0^4}{4} + C 4=404​+C 4=0+C 4 = 0 + C 4=0+C C=4 C = 4 C=4

STEP 5

Substitute the value of C C C back into the general solution to find the particular solution:y=x44+4 y = \frac{x^4}{4} + 4 y=4x4​+4This is the solution to the initial value problem. The particular solution is:y=x44+4 y = \frac{x^4}{4} + 4 y=4x4​+4

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QuestionIn Exercises 47-60, solve the initial value problem.
47. $\frac{d y}{d x}=x^{3}, y(0)=4$

1. The differential equation given is separable and can be solved by integration.
2. The initial condition $y(0) = 4$ will be used to find the constant of integration.

1. Integrate the differential equation to find the general solution.
2. Apply the initial condition to determine the constant of integration.
3. Write the particular solution that satisfies the initial condition.

Integrate the differential equation $\frac{d y}{d x} = x^3$ with respect to $x$ :
$\int \frac{d y}{d x} \, dx = \int x^3 \, dx$
This simplifies to:
$y = \frac{x^4}{4} + C$
where $C$ is the constant of integration.

Apply the initial condition $y(0) = 4$ to find the constant $C$ :
Substitute $x = 0$ and $y = 4$ into the general solution:
$4 = \frac{0^4}{4} + C$ $4 = 0 + C$ $C = 4$

Substitute the value of $C$ back into the general solution to find the particular solution:
$y = \frac{x^4}{4} + 4$
This is the solution to the initial value problem.
The particular solution is:
$y = \frac{x^4}{4} + 4$