Solved on Feb 27, 2024

Solve the ODE $(6+t) \frac{d u}{d t} + u = 6+t$ for $t > 0$ .

STEP 1

Assumptions
1. We are given a first-order linear differential equation of the form: $(6+t) \frac{d u}{d t} + u = 6 + t$
2. The variable $t$ is the independent variable, and $u(t)$ is the dependent variable.
3. The domain of the solution is restricted to $t > 0$ .

STEP 2

The given differential equation is linear and can be written in the standard form: $\frac{d u}{d t} + P(t)u = Q(t)$ where $P(t)$ and $Q(t)$ are functions of $t$ .

STEP 3

Identify $P(t)$ and $Q(t)$ from the given differential equation: $P(t) = \frac{1}{6+t}$ $Q(t) = \frac{6+t}{6+t} = 1$

STEP 4

To solve the linear differential equation, we will use an integrating factor, which is defined as: $\mu(t) = e^{\int P(t) dt}$

STEP 5

Calculate the integrating factor $\mu(t)$ by integrating $P(t)$ : $\mu(t) = e^{\int \frac{1}{6+t} dt}$

STEP 6

Integrate $\frac{1}{6+t}$ with respect to $t$ : $\int \frac{1}{6+t} dt = \ln|6+t|$

STEP 7

Since $t > 0$ , we can ignore the absolute value in the logarithm: $\mu(t) = e^{\ln(6+t)}$

STEP 8

Simplify the integrating factor using the property $e^{\ln(x)} = x$ : $\mu(t) = 6+t$

STEP 9

Multiply the entire differential equation by the integrating factor $\mu(t)$ : $(6+t) \left( (6+t) \frac{d u}{d t} + u \right) = (6+t)(6+t)$

STEP 10

Notice that the left-hand side of the equation is now the derivative of the product of $\mu(t)$ and $u(t)$ : $\frac{d}{dt}[(6+t)u] = (6+t)(6+t)$

STEP 11

Integrate both sides with respect to $t$ : $\int \frac{d}{dt}[(6+t)u] dt = \int (6+t)(6+t) dt$

STEP 12

The left-hand side integrates to $(6+t)u$ : $(6+t)u = \int (6+t)(6+t) dt$

STEP 13

Expand the right-hand side and integrate term by term: $(6+t)u = \int (36 + 12t + t^2) dt$

STEP 14

Integrate each term on the right-hand side: $(6+t)u = 36t + 6t^2/2 + t^3/3 + C$ where $C$ is the constant of integration.

STEP 15

Simplify the integrated terms: $(6+t)u = 36t + 3t^2 + t^3/3 + C$

STEP 16

Divide both sides by $(6+t)$ to solve for $u(t)$ : $u(t) = \frac{36t + 3t^2 + t^3/3 + C}{6+t}$

STEP 17

The solution to the differential equation is: $u(t) = \frac{36t + 3t^2 + t^3/3 + C}{6+t}$ where $C$ is an arbitrary constant that can be determined if an initial condition is given.
Since no initial condition is provided, the solution contains an arbitrary constant $C$ . This is the general solution to the differential equation.

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Solve the ODE $(6+t) \frac{d u}{d t} + u = 6+t$ for $t > 0$ .

STEP 1

Assumptions
1. We are given a first-order linear differential equation of the form: $(6+t) \frac{d u}{d t} + u = 6 + t$
2. The variable $t$ is the independent variable, and $u(t)$ is the dependent variable.
3. The domain of the solution is restricted to $t > 0$ .

STEP 2

The given differential equation is linear and can be written in the standard form: $\frac{d u}{d t} + P(t)u = Q(t)$ where $P(t)$ and $Q(t)$ are functions of $t$ .

STEP 3

Identify $P(t)$ and $Q(t)$ from the given differential equation: $P(t) = \frac{1}{6+t}$ $Q(t) = \frac{6+t}{6+t} = 1$

STEP 4

To solve the linear differential equation, we will use an integrating factor, which is defined as: $\mu(t) = e^{\int P(t) dt}$

STEP 5

Calculate the integrating factor $\mu(t)$ by integrating $P(t)$ : $\mu(t) = e^{\int \frac{1}{6+t} dt}$

STEP 6

Integrate $\frac{1}{6+t}$ with respect to $t$ : $\int \frac{1}{6+t} dt = \ln|6+t|$

STEP 7

Since $t > 0$ , we can ignore the absolute value in the logarithm: $\mu(t) = e^{\ln(6+t)}$

STEP 8

Simplify the integrating factor using the property $e^{\ln(x)} = x$ : $\mu(t) = 6+t$

STEP 9

Multiply the entire differential equation by the integrating factor $\mu(t)$ : $(6+t) \left( (6+t) \frac{d u}{d t} + u \right) = (6+t)(6+t)$

STEP 10

Notice that the left-hand side of the equation is now the derivative of the product of $\mu(t)$ and $u(t)$ : $\frac{d}{dt}[(6+t)u] = (6+t)(6+t)$

STEP 11

Integrate both sides with respect to $t$ : $\int \frac{d}{dt}[(6+t)u] dt = \int (6+t)(6+t) dt$

STEP 12

The left-hand side integrates to $(6+t)u$ : $(6+t)u = \int (6+t)(6+t) dt$

STEP 13

Expand the right-hand side and integrate term by term: $(6+t)u = \int (36 + 12t + t^2) dt$

STEP 14

Integrate each term on the right-hand side: $(6+t)u = 36t + 6t^2/2 + t^3/3 + C$ where $C$ is the constant of integration.

STEP 15

Simplify the integrated terms: $(6+t)u = 36t + 3t^2 + t^3/3 + C$

STEP 16

Divide both sides by $(6+t)$ to solve for $u(t)$ : $u(t) = \frac{36t + 3t^2 + t^3/3 + C}{6+t}$

STEP 17

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Solve the ODE (6+t)dudt+u=6+t(6+t) \frac{d u}{d t} + u = 6+t(6+t)dtdu​+u=6+t for t>0t > 0t>0.

STEP 1

STEP 2

The given differential equation is linear and can be written in the standard form: dudt+P(t)u=Q(t) \frac{d u}{d t} + P(t)u = Q(t) dtdu​+P(t)u=Q(t) where P(t)P(t)P(t) and Q(t)Q(t)Q(t) are functions of ttt.

STEP 3

Identify P(t)P(t)P(t) and Q(t)Q(t)Q(t) from the given differential equation: P(t)=16+t P(t) = \frac{1}{6+t} P(t)=6+t1​ Q(t)=6+t6+t=1 Q(t) = \frac{6+t}{6+t} = 1 Q(t)=6+t6+t​=1

STEP 4

To solve the linear differential equation, we will use an integrating factor, which is defined as: μ(t)=e∫P(t)dt \mu(t) = e^{\int P(t) dt} μ(t)=e∫P(t)dt

STEP 5

Calculate the integrating factor μ(t)\mu(t)μ(t) by integrating P(t)P(t)P(t): μ(t)=e∫16+tdt \mu(t) = e^{\int \frac{1}{6+t} dt} μ(t)=e∫6+t1​dt

STEP 6

Integrate 16+t\frac{1}{6+t}6+t1​ with respect to ttt: ∫16+tdt=ln⁡∣6+t∣ \int \frac{1}{6+t} dt = \ln|6+t| ∫6+t1​dt=ln∣6+t∣

STEP 7

Since t>0t > 0t>0, we can ignore the absolute value in the logarithm: μ(t)=eln⁡(6+t) \mu(t) = e^{\ln(6+t)} μ(t)=eln(6+t)

STEP 8

Simplify the integrating factor using the property eln⁡(x)=xe^{\ln(x)} = xeln(x)=x: μ(t)=6+t \mu(t) = 6+t μ(t)=6+t

STEP 9

Multiply the entire differential equation by the integrating factor μ(t)\mu(t)μ(t): (6+t)((6+t)dudt+u)=(6+t)(6+t) (6+t) \left( (6+t) \frac{d u}{d t} + u \right) = (6+t)(6+t) (6+t)((6+t)dtdu​+u)=(6+t)(6+t)

STEP 10

Notice that the left-hand side of the equation is now the derivative of the product of μ(t)\mu(t)μ(t) and u(t)u(t)u(t): ddt[(6+t)u]=(6+t)(6+t) \frac{d}{dt}[(6+t)u] = (6+t)(6+t) dtd​[(6+t)u]=(6+t)(6+t)

STEP 11

Integrate both sides with respect to ttt: ∫ddt[(6+t)u]dt=∫(6+t)(6+t)dt \int \frac{d}{dt}[(6+t)u] dt = \int (6+t)(6+t) dt ∫dtd​[(6+t)u]dt=∫(6+t)(6+t)dt

STEP 12

The left-hand side integrates to (6+t)u(6+t)u(6+t)u: (6+t)u=∫(6+t)(6+t)dt (6+t)u = \int (6+t)(6+t) dt (6+t)u=∫(6+t)(6+t)dt

STEP 13

Expand the right-hand side and integrate term by term: (6+t)u=∫(36+12t+t2)dt (6+t)u = \int (36 + 12t + t^2) dt (6+t)u=∫(36+12t+t2)dt

STEP 14

Integrate each term on the right-hand side: (6+t)u=36t+6t2/2+t3/3+C (6+t)u = 36t + 6t^2/2 + t^3/3 + C (6+t)u=36t+6t2/2+t3/3+C where CCC is the constant of integration.

STEP 15

Simplify the integrated terms: (6+t)u=36t+3t2+t3/3+C (6+t)u = 36t + 3t^2 + t^3/3 + C (6+t)u=36t+3t2+t3/3+C

STEP 16

Divide both sides by (6+t)(6+t)(6+t) to solve for u(t)u(t)u(t): u(t)=36t+3t2+t3/3+C6+t u(t) = \frac{36t + 3t^2 + t^3/3 + C}{6+t} u(t)=6+t36t+3t2+t3/3+C​

STEP 17

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Solve the ODE $(6+t) \frac{d u}{d t} + u = 6+t$ for $t > 0$ .

The given differential equation is linear and can be written in the standard form: $\frac{d u}{d t} + P(t)u = Q(t)$ where $P(t)$ and $Q(t)$ are functions of $t$ .

Identify $P(t)$ and $Q(t)$ from the given differential equation: $P(t) = \frac{1}{6+t}$ $Q(t) = \frac{6+t}{6+t} = 1$

To solve the linear differential equation, we will use an integrating factor, which is defined as: $\mu(t) = e^{\int P(t) dt}$

Calculate the integrating factor $\mu(t)$ by integrating $P(t)$ : $\mu(t) = e^{\int \frac{1}{6+t} dt}$

Integrate $\frac{1}{6+t}$ with respect to $t$ : $\int \frac{1}{6+t} dt = \ln|6+t|$

Since $t > 0$ , we can ignore the absolute value in the logarithm: $\mu(t) = e^{\ln(6+t)}$

Simplify the integrating factor using the property $e^{\ln(x)} = x$ : $\mu(t) = 6+t$

Multiply the entire differential equation by the integrating factor $\mu(t)$ : $(6+t) \left( (6+t) \frac{d u}{d t} + u \right) = (6+t)(6+t)$

Notice that the left-hand side of the equation is now the derivative of the product of $\mu(t)$ and $u(t)$ : $\frac{d}{dt}[(6+t)u] = (6+t)(6+t)$

Integrate both sides with respect to $t$ : $\int \frac{d}{dt}[(6+t)u] dt = \int (6+t)(6+t) dt$

The left-hand side integrates to $(6+t)u$ : $(6+t)u = \int (6+t)(6+t) dt$

Expand the right-hand side and integrate term by term: $(6+t)u = \int (36 + 12t + t^2) dt$

Integrate each term on the right-hand side: $(6+t)u = 36t + 6t^2/2 + t^3/3 + C$ where $C$ is the constant of integration.

Simplify the integrated terms: $(6+t)u = 36t + 3t^2 + t^3/3 + C$

Divide both sides by $(6+t)$ to solve for $u(t)$ : $u(t) = \frac{36t + 3t^2 + t^3/3 + C}{6+t}$