Solved on Dec 10, 2023

Solve the initial value problem $\frac{dv}{dt}=\frac{2}{t\sqrt{t^2-1}}, t>1, v(2)=0$ and find $v(t)$ .

STEP 1

Assumptions
1. We are given the differential equation $\frac{dv}{dt} = \frac{2}{t\sqrt{t^2 - 1}}$ .
2. The domain of the function is $t > 1$ .
3. The initial condition is $v(2) = 0$ .
4. We are to solve for $v(t)$ .

STEP 2

To solve the initial value problem, we need to integrate the right-hand side of the differential equation with respect to $t$ .
$v(t) = \int \frac{2}{t\sqrt{t^2 - 1}} \, dt$

STEP 3

To integrate the function, we will perform a substitution. Let $u = t^2 - 1$ . Then, $du = 2t \, dt$ .

STEP 4

We rewrite the integral in terms of $u$ using the substitution from STEP_3.
$v(t) = \int \frac{1}{\sqrt{u}} \, du$

STEP 5

Now we integrate the function with respect to $u$ .
$v(t) = \int u^{-\frac{1}{2}} \, du$

STEP 6

The integral of $u^{-\frac{1}{2}}$ with respect to $u$ is $2u^{\frac{1}{2}}$ .
$v(t) = 2u^{\frac{1}{2}} + C$

STEP 7

We now substitute back for $u$ using $u = t^2 - 1$ to get $v(t)$ in terms of $t$ .
$v(t) = 2(t^2 - 1)^{\frac{1}{2}} + C$

STEP 8

We apply the initial condition $v(2) = 0$ to find the constant $C$ .
$0 = 2(2^2 - 1)^{\frac{1}{2}} + C$

STEP 9

Calculate the term $(2^2 - 1)^{\frac{1}{2}}$ .
$0 = 2(3)^{\frac{1}{2}} + C$

STEP 10

Solve for $C$ .
$C = -2(3)^{\frac{1}{2}}$

STEP 11

Substitute the value of $C$ back into the expression for $v(t)$ .
$v(t) = 2(t^2 - 1)^{\frac{1}{2}} - 2(3)^{\frac{1}{2}}$

STEP 12

Simplify the expression for $v(t)$ .
$v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}$
This is the solution to the initial value problem.
$v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}$

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Solve the initial value problem $\frac{dv}{dt}=\frac{2}{t\sqrt{t^2-1}}, t>1, v(2)=0$ and find $v(t)$ .

STEP 1

Assumptions
1. We are given the differential equation $\frac{dv}{dt} = \frac{2}{t\sqrt{t^2 - 1}}$ .
2. The domain of the function is $t > 1$ .
3. The initial condition is $v(2) = 0$ .
4. We are to solve for $v(t)$ .

STEP 2

To solve the initial value problem, we need to integrate the right-hand side of the differential equation with respect to $t$ .
$v(t) = \int \frac{2}{t\sqrt{t^2 - 1}} \, dt$

STEP 3

To integrate the function, we will perform a substitution. Let $u = t^2 - 1$ . Then, $du = 2t \, dt$ .

STEP 4

We rewrite the integral in terms of $u$ using the substitution from STEP_3.
$v(t) = \int \frac{1}{\sqrt{u}} \, du$

STEP 5

Now we integrate the function with respect to $u$ .
$v(t) = \int u^{-\frac{1}{2}} \, du$

STEP 6

The integral of $u^{-\frac{1}{2}}$ with respect to $u$ is $2u^{\frac{1}{2}}$ .
$v(t) = 2u^{\frac{1}{2}} + C$

STEP 7

We now substitute back for $u$ using $u = t^2 - 1$ to get $v(t)$ in terms of $t$ .
$v(t) = 2(t^2 - 1)^{\frac{1}{2}} + C$

STEP 8

We apply the initial condition $v(2) = 0$ to find the constant $C$ .
$0 = 2(2^2 - 1)^{\frac{1}{2}} + C$

STEP 9

Calculate the term $(2^2 - 1)^{\frac{1}{2}}$ .
$0 = 2(3)^{\frac{1}{2}} + C$

STEP 10

Solve for $C$ .
$C = -2(3)^{\frac{1}{2}}$

STEP 11

Substitute the value of $C$ back into the expression for $v(t)$ .
$v(t) = 2(t^2 - 1)^{\frac{1}{2}} - 2(3)^{\frac{1}{2}}$

STEP 12

Simplify the expression for $v(t)$ .
$v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}$
This is the solution to the initial value problem.
$v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}$

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Solve the initial value problem dvdt=2tt2−1,t>1,v(2)=0\frac{dv}{dt}=\frac{2}{t\sqrt{t^2-1}}, t>1, v(2)=0dtdv​=tt2−1​2​,t>1,v(2)=0 and find v(t)v(t)v(t).

STEP 1

Assumptions1. We are given the differential equation dvdt=2tt2−1\frac{dv}{dt} = \frac{2}{t\sqrt{t^2 - 1}}dtdv​=tt2−1​2​.2. The domain of the function is t>1t > 1t>1.3. The initial condition is v(2)=0v(2) = 0v(2)=0.4. We are to solve for v(t)v(t)v(t).

STEP 2

To solve the initial value problem, we need to integrate the right-hand side of the differential equation with respect to ttt.v(t)=∫2tt2−1 dtv(t) = \int \frac{2}{t\sqrt{t^2 - 1}} \, dtv(t)=∫tt2−1​2​dt

STEP 3

To integrate the function, we will perform a substitution. Let u=t2−1u = t^2 - 1u=t2−1. Then, du=2t dtdu = 2t \, dtdu=2tdt.

STEP 4

We rewrite the integral in terms of uuu using the substitution from STEP_3.v(t)=∫1u duv(t) = \int \frac{1}{\sqrt{u}} \, duv(t)=∫u​1​du

STEP 5

Now we integrate the function with respect to uuu.v(t)=∫u−12 duv(t) = \int u^{-\frac{1}{2}} \, duv(t)=∫u−21​du

STEP 6

The integral of u−12u^{-\frac{1}{2}}u−21​ with respect to uuu is 2u122u^{\frac{1}{2}}2u21​.v(t)=2u12+Cv(t) = 2u^{\frac{1}{2}} + Cv(t)=2u21​+C

STEP 7

We now substitute back for uuu using u=t2−1u = t^2 - 1u=t2−1 to get v(t)v(t)v(t) in terms of ttt.v(t)=2(t2−1)12+Cv(t) = 2(t^2 - 1)^{\frac{1}{2}} + Cv(t)=2(t2−1)21​+C

STEP 8

We apply the initial condition v(2)=0v(2) = 0v(2)=0 to find the constant CCC.0=2(22−1)12+C0 = 2(2^2 - 1)^{\frac{1}{2}} + C0=2(22−1)21​+C

STEP 9

Calculate the term (22−1)12(2^2 - 1)^{\frac{1}{2}}(22−1)21​.0=2(3)12+C0 = 2(3)^{\frac{1}{2}} + C0=2(3)21​+C

STEP 10

Solve for CCC.C=−2(3)12C = -2(3)^{\frac{1}{2}}C=−2(3)21​

STEP 11

Substitute the value of CCC back into the expression for v(t)v(t)v(t).v(t)=2(t2−1)12−2(3)12v(t) = 2(t^2 - 1)^{\frac{1}{2}} - 2(3)^{\frac{1}{2}}v(t)=2(t2−1)21​−2(3)21​

STEP 12

Simplify the expression for v(t)v(t)v(t).v(t)=2t2−1−23v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}v(t)=2t2−1​−23​This is the solution to the initial value problem.v(t)=2t2−1−23v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}v(t)=2t2−1​−23​

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Solve the initial value problem $\frac{dv}{dt}=\frac{2}{t\sqrt{t^2-1}}, t>1, v(2)=0$ and find $v(t)$ .

Assumptions
1. We are given the differential equation $\frac{dv}{dt} = \frac{2}{t\sqrt{t^2 - 1}}$ .
2. The domain of the function is $t > 1$ .
3. The initial condition is $v(2) = 0$ .
4. We are to solve for $v(t)$ .

To solve the initial value problem, we need to integrate the right-hand side of the differential equation with respect to $t$ .
$v(t) = \int \frac{2}{t\sqrt{t^2 - 1}} \, dt$

To integrate the function, we will perform a substitution. Let $u = t^2 - 1$ . Then, $du = 2t \, dt$ .

We rewrite the integral in terms of $u$ using the substitution from STEP_3.
$v(t) = \int \frac{1}{\sqrt{u}} \, du$

Now we integrate the function with respect to $u$ .
$v(t) = \int u^{-\frac{1}{2}} \, du$

The integral of $u^{-\frac{1}{2}}$ with respect to $u$ is $2u^{\frac{1}{2}}$ .
$v(t) = 2u^{\frac{1}{2}} + C$

We now substitute back for $u$ using $u = t^2 - 1$ to get $v(t)$ in terms of $t$ .
$v(t) = 2(t^2 - 1)^{\frac{1}{2}} + C$

We apply the initial condition $v(2) = 0$ to find the constant $C$ .
$0 = 2(2^2 - 1)^{\frac{1}{2}} + C$

Calculate the term $(2^2 - 1)^{\frac{1}{2}}$ .
$0 = 2(3)^{\frac{1}{2}} + C$

Solve for $C$ .
$C = -2(3)^{\frac{1}{2}}$

Substitute the value of $C$ back into the expression for $v(t)$ .
$v(t) = 2(t^2 - 1)^{\frac{1}{2}} - 2(3)^{\frac{1}{2}}$

Simplify the expression for $v(t)$ .
$v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}$
This is the solution to the initial value problem.
$v(t) = 2\sqrt{t^2 - 1} - 2\sqrt{3}$