Math / Calculus

QuestionSolve the separable differential equation $\frac{d y}{d x}=-8 y$ , and find the particular solution satisfying the initial condition $y(0)=9$ . $y(x)=$ $\square$

Studdy Solution

STEP 1

1. The differential equation $\frac{dy}{dx} = -8y$ is separable.
2. We will use separation of variables to solve the differential equation.
3. The initial condition $y(0) = 9$ will be used to find the particular solution.

STEP 2

1. Separate the variables.
2. Integrate both sides.
3. Solve for the constant of integration using the initial condition.
4. Write the particular solution.

STEP 3

Separate the variables by dividing both sides by $y$ and multiplying both sides by $dx$ :
$\frac{1}{y} \, dy = -8 \, dx$

STEP 4

Integrate both sides:
$\int \frac{1}{y} \, dy = \int -8 \, dx$
This gives:
$\ln |y| = -8x + C$
where $C$ is the constant of integration.

STEP 5

Solve for $y$ by exponentiating both sides to eliminate the natural logarithm:
$|y| = e^{-8x + C}$
We can rewrite this as:
$y = \pm e^C \cdot e^{-8x}$
Let $C_1 = \pm e^C$ , then:
$y = C_1 e^{-8x}$
Use the initial condition $y(0) = 9$ to solve for $C_1$ :
$9 = C_1 e^{0}$ $C_1 = 9$

STEP 6

Substitute $C_1$ back into the equation for $y$ :
$y = 9 e^{-8x}$
The particular solution satisfying the initial condition is:
$y(x) = 9 e^{-8x}$

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QuestionSolve the separable differential equation $\frac{d y}{d x}=-8 y$ , and find the particular solution satisfying the initial condition $y(0)=9$ . $y(x)=$ $\square$

Studdy Solution

STEP 1

1. The differential equation $\frac{dy}{dx} = -8y$ is separable.
2. We will use separation of variables to solve the differential equation.
3. The initial condition $y(0) = 9$ will be used to find the particular solution.

STEP 2

1. Separate the variables.
2. Integrate both sides.
3. Solve for the constant of integration using the initial condition.
4. Write the particular solution.

STEP 3

Separate the variables by dividing both sides by $y$ and multiplying both sides by $dx$ :
$\frac{1}{y} \, dy = -8 \, dx$

STEP 4

Integrate both sides:
$\int \frac{1}{y} \, dy = \int -8 \, dx$
This gives:
$\ln |y| = -8x + C$
where $C$ is the constant of integration.

STEP 5

Solve for $y$ by exponentiating both sides to eliminate the natural logarithm:
$|y| = e^{-8x + C}$
We can rewrite this as:
$y = \pm e^C \cdot e^{-8x}$
Let $C_1 = \pm e^C$ , then:
$y = C_1 e^{-8x}$
Use the initial condition $y(0) = 9$ to solve for $C_1$ :
$9 = C_1 e^{0}$ $C_1 = 9$

STEP 6

Substitute $C_1$ back into the equation for $y$ :
$y = 9 e^{-8x}$
The particular solution satisfying the initial condition is:
$y(x) = 9 e^{-8x}$

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QuestionSolve the separable differential equation dydx=−8y\frac{d y}{d x}=-8 ydxdy​=−8y, and find the particular solution satisfying the initial condition y(0)=9y(0)=9y(0)=9. y(x)=y(x)=y(x)= □\square□

Studdy Solution

STEP 1

1. The differential equation dydx=−8y\frac{dy}{dx} = -8ydxdy​=−8y is separable.2. We will use separation of variables to solve the differential equation.3. The initial condition y(0)=9y(0) = 9y(0)=9 will be used to find the particular solution.

STEP 2

1. Separate the variables.2. Integrate both sides.3. Solve for the constant of integration using the initial condition.4. Write the particular solution.

STEP 3

Separate the variables by dividing both sides by yyy and multiplying both sides by dxdxdx:1y dy=−8 dx\frac{1}{y} \, dy = -8 \, dxy1​dy=−8dx

STEP 4

Integrate both sides:∫1y dy=∫−8 dx\int \frac{1}{y} \, dy = \int -8 \, dx∫y1​dy=∫−8dxThis gives:ln⁡∣y∣=−8x+C\ln |y| = -8x + Cln∣y∣=−8x+Cwhere CCC is the constant of integration.

STEP 5

STEP 6

Substitute C1C_1C1​ back into the equation for yyy:y=9e−8xy = 9 e^{-8x}y=9e−8x The particular solution satisfying the initial condition is:y(x)=9e−8xy(x) = 9 e^{-8x}y(x)=9e−8x

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QuestionSolve the separable differential equation $\frac{d y}{d x}=-8 y$ , and find the particular solution satisfying the initial condition $y(0)=9$ . $y(x)=$ $\square$

1. The differential equation $\frac{dy}{dx} = -8y$ is separable.
2. We will use separation of variables to solve the differential equation.
3. The initial condition $y(0) = 9$ will be used to find the particular solution.

1. Separate the variables.
2. Integrate both sides.
3. Solve for the constant of integration using the initial condition.
4. Write the particular solution.

Separate the variables by dividing both sides by $y$ and multiplying both sides by $dx$ :
$\frac{1}{y} \, dy = -8 \, dx$

Integrate both sides:
$\int \frac{1}{y} \, dy = \int -8 \, dx$
This gives:
$\ln |y| = -8x + C$
where $C$ is the constant of integration.

Substitute $C_1$ back into the equation for $y$ :
$y = 9 e^{-8x}$
The particular solution satisfying the initial condition is:
$y(x) = 9 e^{-8x}$