Math / Calculus

Question(12) Let $y_{1}$ and $y_{2}$ be two solutions of the DE $t^{2} y^{\prime \prime}-t(t+1) y^{\prime}+y=0, \quad t>0 .$
If $W\left(y_{1}, y_{2}\right)(2)=2 \mathrm{e}^{2}$ , then $W\left(y_{1}, y_{2}\right)(-1)=$ (a) -e (b) $\mathrm{e}^{-1}$ (c) $e^{2}$ (d) $-\mathrm{e}^{-1}$ (e) $e$

Studdy Solution

STEP 1

1. The differential equation given is a second-order linear homogeneous differential equation.
2. The Wronskian $W(y_1, y_2)$ is a function of $t$ that can be used to determine the linear independence of the solutions $y_1$ and $y_2$ .
3. The Wronskian satisfies a first-order linear differential equation derived from the original differential equation.

STEP 2

1. Recall the formula for the Wronskian of solutions to a second-order linear differential equation.
2. Use the differential equation to find the expression for the derivative of the Wronskian.
3. Solve the differential equation for the Wronskian.
4. Use the given condition to find the constant of integration.
5. Evaluate the Wronskian at $t = -1$ .

STEP 3

Recall the formula for the Wronskian of solutions to a second-order linear differential equation:
For a second-order linear differential equation of the form $y'' + p(t)y' + q(t)y = 0$ , the Wronskian $W(y_1, y_2)$ satisfies:
$W'(t) = -p(t)W(t)$

STEP 4

Identify $p(t)$ from the given differential equation:
The given differential equation is:
$t^2 y'' - t(t+1) y' + y = 0$
Rewriting it in standard form:
$y'' - \frac{t+1}{t} y' + \frac{1}{t^2} y = 0$
Thus, $p(t) = -\frac{t+1}{t}$ .

STEP 5

Write the differential equation for the Wronskian:
$W'(t) = -\left(-\frac{t+1}{t}\right) W(t) = \frac{t+1}{t} W(t)$
This is a separable differential equation.

STEP 6

Solve the differential equation for $W(t)$ :
Separate variables:
$\frac{dW}{W} = \frac{t+1}{t} dt$
Integrate both sides:
$\int \frac{dW}{W} = \int \left(1 + \frac{1}{t}\right) dt$
$\ln |W| = t + \ln |t| + C$
Exponentiate both sides:
$|W| = e^{t} \cdot |t| \cdot e^C$
Let $C_1 = e^C$ , then:
$W(t) = C_1 e^{t} |t|$

STEP 7

Use the given condition $W(2) = 2e^2$ to find $C_1$ :
$W(2) = C_1 e^{2} \cdot 2 = 2e^2$
$C_1 \cdot 2e^2 = 2e^2$
$C_1 = 1$
Thus, the Wronskian is:
$W(t) = e^{t} |t|$

STEP 8

Evaluate the Wronskian at $t = -1$ :
$W(-1) = e^{-1} \cdot |-1| = e^{-1}$
The value of the Wronskian at $t = -1$ is $\boxed{e^{-1}}$ .

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Question(12) Let $y_{1}$ and $y_{2}$ be two solutions of the DE $t^{2} y^{\prime \prime}-t(t+1) y^{\prime}+y=0, \quad t>0 .$
If $W\left(y_{1}, y_{2}\right)(2)=2 \mathrm{e}^{2}$ , then $W\left(y_{1}, y_{2}\right)(-1)=$ (a) -e (b) $\mathrm{e}^{-1}$ (c) $e^{2}$ (d) $-\mathrm{e}^{-1}$ (e) $e$

Studdy Solution

STEP 1

STEP 2

STEP 3

Recall the formula for the Wronskian of solutions to a second-order linear differential equation:
For a second-order linear differential equation of the form $y'' + p(t)y' + q(t)y = 0$ , the Wronskian $W(y_1, y_2)$ satisfies:
$W'(t) = -p(t)W(t)$

STEP 4

Identify $p(t)$ from the given differential equation:
The given differential equation is:
$t^2 y'' - t(t+1) y' + y = 0$
Rewriting it in standard form:
$y'' - \frac{t+1}{t} y' + \frac{1}{t^2} y = 0$
Thus, $p(t) = -\frac{t+1}{t}$ .

STEP 5

Write the differential equation for the Wronskian:
$W'(t) = -\left(-\frac{t+1}{t}\right) W(t) = \frac{t+1}{t} W(t)$
This is a separable differential equation.

STEP 6

STEP 7

Use the given condition $W(2) = 2e^2$ to find $C_1$ :
$W(2) = C_1 e^{2} \cdot 2 = 2e^2$
$C_1 \cdot 2e^2 = 2e^2$
$C_1 = 1$
Thus, the Wronskian is:
$W(t) = e^{t} |t|$

STEP 8

Evaluate the Wronskian at $t = -1$ :
$W(-1) = e^{-1} \cdot |-1| = e^{-1}$
The value of the Wronskian at $t = -1$ is $\boxed{e^{-1}}$ .

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Studdy Solution

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

Write the differential equation for the Wronskian:W′(t)=−(−t+1t)W(t)=t+1tW(t) W'(t) = -\left(-\frac{t+1}{t}\right) W(t) = \frac{t+1}{t} W(t) W′(t)=−(−tt+1​)W(t)=tt+1​W(t)This is a separable differential equation.

STEP 6

STEP 7

Use the given condition W(2)=2e2 W(2) = 2e^2 W(2)=2e2 to find C1 C_1 C1​:W(2)=C1e2⋅2=2e2 W(2) = C_1 e^{2} \cdot 2 = 2e^2 W(2)=C1​e2⋅2=2e2C1⋅2e2=2e2 C_1 \cdot 2e^2 = 2e^2 C1​⋅2e2=2e2C1=1 C_1 = 1 C1​=1Thus, the Wronskian is:W(t)=et∣t∣ W(t) = e^{t} |t| W(t)=et∣t∣

STEP 8

Evaluate the Wronskian at t=−1 t = -1 t=−1:W(−1)=e−1⋅∣−1∣=e−1 W(-1) = e^{-1} \cdot |-1| = e^{-1} W(−1)=e−1⋅∣−1∣=e−1 The value of the Wronskian at t=−1 t = -1 t=−1 is e−1 \boxed{e^{-1}} e−1​.

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Write the differential equation for the Wronskian:
$W'(t) = -\left(-\frac{t+1}{t}\right) W(t) = \frac{t+1}{t} W(t)$
This is a separable differential equation.

Use the given condition $W(2) = 2e^2$ to find $C_1$ :
$W(2) = C_1 e^{2} \cdot 2 = 2e^2$
$C_1 \cdot 2e^2 = 2e^2$
$C_1 = 1$
Thus, the Wronskian is:
$W(t) = e^{t} |t|$

Evaluate the Wronskian at $t = -1$ :
$W(-1) = e^{-1} \cdot |-1| = e^{-1}$
The value of the Wronskian at $t = -1$ is $\boxed{e^{-1}}$ .