To solve the given initial value problem: \[ y'' - \frac{2}{7} y' + \frac{y}{49} = 0, \quad y(0) = 30, \quad y'(0) = b, \] we start by solving the characteristic equation associated with the differential equation. The characteristic equation is: \[ r^2 - \frac{2}{7}r + \frac{1}{49} = 0. \] This is a quadratic equation, and we can solve it using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 1 \), \( b = -\frac{2}{7} \), and \( c = \frac{1}{49} \). Substituting these values in, we get: \[ r = \frac{\frac{2}{7} \pm \sqrt{\left(\frac{2}{7}\right)^2 - 4 \cdot 1 \cdot \frac{1}{49}}}{2 \cdot 1} = \frac{\frac{2}{7} \pm \sqrt{\frac{4}{49} - \frac{4}{49}}}{2} = \frac{\frac{2}{7} \pm \sqrt{0}}{2} = \frac{\frac{2}{7}}{2} = \frac{1}{7}. \] So, we have a repeated root \( r = \frac{1}{7} \). For a repeated root, the general solution to the differential equation is: \[ y(t) = (C_1 + C_2 t) e^{rt} = (C_1 + C_2 t) e^{\frac{t}{7}}. \] Next, we use the initial conditions to find \( C_1 \) and \( C_2 \). 1. Using \( y(0) = 30 \): \[ y(0) = (C_1 + C_2 \cdot 0) e^{0} = C_1 = 30. \] So, \( C_1 = 30 \). 2. Using \( y'(0) = b \): First, we find \( y'(t) \): \[ y'(t) = \left( C_2 e^{\frac{t}{7}} + (C_1 + C_2 t) \cdot \frac{1}{7} e^{\frac{t}{7}} \right) = e^{\frac{t}{7}} \left( C_2 + \frac{1}{7} (C_1 + C_2 t) \right). \] Evaluating at \( t = 0 \): \[ y'(0) = e^{0} \left( C_2 + \frac{1}{7} C_1 \right) = C_2 + \frac{1}{7} \cdot 30 = C_2 + \frac{30}{7} = b. \] So, \[ C_2 = b - \frac{30}{7}. \] Thus, the solution is: \[ y(t) = \left( 30 + \left( b - \frac{30}{7} \right) t \right) e^{\frac{t}{7}}. \] To determine the critical value of \( b \) that separates solutions that remain positive for all \( t > 0 \) from those that eventually become negative, we need to ensure that the term inside the parentheses does not become negative for any \( t \geq 0 \): \[ 30 + \left( b - \frac{30}{7} \right) t > 0 \quad \text{for all} \quad t \geq 0. \] For \( t = 0 \), this is always positive since \( 30 > 0 \). For large \( t \), the term \( \left( b - \frac{30}{7} \right) t \) must be non-negative: \[ b - \frac{30}{7} \geq 0 \implies b \geq \frac{30}{7}. \] So, the critical value of \( b \) is: \[ \boxed{\frac{30}{7}}. \]

To solve the given initial value problem: \[ y' - y = 7te^{2t}, \quad y(0) = 1 \] we can use the method of integrating factors. The differential equation is a first-order linear differential equation of the form: \[ y' - y = 7te^{2t} \] First, we identify the integrating factor \( \mu(t) \): \[ \mu(t) = e^{\int -1 \, dt} = e^{-t} \] Next, we multiply both sides of the differential equation by the integrating factor: \[ e^{-t} y' - e^{-t} y = 7te^{2t} e^{-t} \] \[ e^{-t} y' - e^{-t} y = 7t e^{t} \] The left-hand side of the equation is the derivative of \( e^{-t} y \): \[ \frac{d}{dt} (e^{-t} y) = 7t e^{t} \] Now, we integrate both sides with respect to \( t \): \[ \int \frac{d}{dt} (e^{-t} y) \, dt = \int 7t e^{t} \, dt \] The left-hand side integrates to: \[ e^{-t} y \] For the right-hand side, we use integration by parts. Let: \[ u = t \quad \text{and} \quad dv = 7e^{t} \, dt \] \[ du = dt \quad \text{and} \quad v = 7e^{t} \] Then: \[ \int 7t e^{t} \, dt = 7t e^{t} - \int 7e^{t} \, dt \] \[ = 7t e^{t} - 7e^{t} \] \[ = 7e^{t} (t - 1) \] So, we have: \[ e^{-t} y = 7e^{t} (t - 1) + C \] Multiplying both sides by \( e^{t} \): \[ y = 7e^{t} (t - 1) e^{t} + C e^{t} \] \[ y = 7e^{2t} (t - 1) + C e^{t} \] Now, we use the initial condition \( y(0) = 1 \): \[ 1 = 7e^{0} (0 - 1) + C e^{0} \] \[ 1 = -7 + C \] \[ C = 8 \] Therefore, the solution to the initial value problem is: \[ y(t) = 7e^{2t} (t - 1) + 8e^{t} \] So, the final solution is: \[ y(t) = 7te^{2t} - 7e^{2t} + 8e^{t} \]

To solve the initial value problem \(25y'' - 40y' + 16y = 0\) with \(y(0) = 2\) and \(y'(0) = 3\), we first find the general solution to the differential equation. ### Step 1: Find the characteristic equationThe characteristic equation for the differential equation \(25y'' - 40y' + 16y = 0\) is: \[25r^2 - 40r + 16 = 0\] ### Step 2: Solve the characteristic equationWe solve the quadratic equation using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 25\), \(b = -40\), and \(c = 16\): \[r = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 25 \cdot 16}}{2 \cdot 25}\] \[r = \frac{40 \pm \sqrt{1600 - 1600}}{50}\] \[r = \frac{40 \pm \sqrt{0}}{50}\] \[r = \frac{40}{50}\] \[r = \frac{4}{5}\] Since the discriminant (\(b^2 - 4ac\)) is zero, we have a repeated root \(r = \frac{4}{5}\). For a repeated root, the general solution to the differential equation is: \[y(t) = (c_1 + c_2 t)e^{rt}\] Substituting \(r = \frac{4}{5}\), we get: \[y(t) = (c_1 + c_2 t)e^{\frac{4}{5}t}\] ### Step 3: Apply initial conditionsWe use the initial conditions \(y(0) = 2\) and \(y'(0) = 3\) to find \(c_1\) and \(c_2\). #### Initial condition \(y(0) = 2\): \[y(0) = (c_1 + c_2 \cdot 0)e^{\frac{4}{5} \cdot 0} = c_1 = 2\] So, \(c_1 = 2\). #### Initial condition \(y'(0) = 3\): First, we find \(y'(t)\): \[y(t) = (2 + c_2 t)e^{\frac{4}{5}t}\] \[y'(t) = \left[2 + c_2 t\right] \cdot \frac{4}{5}e^{\frac{4}{5}t} + c_2 e^{\frac{4}{5}t}\] \[y'(t) = \left(\frac{8}{5} + \frac{4}{5}c_2 t + c_2\right)e^{\frac{4}{5}t}\] Now, apply \(y'(0) = 3\): \[y'(0) = \left(\frac{8}{5} + c_2\right)e^{0} = \frac{8}{5} + c_2 = 3\] Solving for \(c_2\): \[\frac{8}{5} + c_2 = 3\] \[c_2 = 3 - \frac{8}{5}\] \[c_2 = \frac{15}{5} - \frac{8}{5}\] \[c_2 = \frac{7}{5}\] ### Step 4: Write the final solutionSubstitute \(c_1\) and \(c_2\) back into the general solution: \[y(t) = \left(2 + \frac{7}{5}t\right)e^{\frac{4}{5}t}\] So, the solution to the initial value problem is: \[y(t) = \left(2 + \frac{7}{5}t\right)e^{\frac{4}{5}t}\]

To solve the initial value problem 42𝑦′′′+13𝑦′′+𝑦′=042y ′′′ +13y ′′ +y ′ =0 with initial conditions 𝑦(0)=−7y(0)=−7, 𝑦′(0)=2y ′ (0)=2, and 𝑦′′(0)=0y ′′ (0)=0, we follow these steps: 1. **Find the characteristic equation:** The given differential equation is:42𝑦′′′+13𝑦′′+𝑦′=042y ′′′ +13y ′′ +y ′ =0Assume a solution of the form 𝑦=𝑒𝑟𝑡y=e rt . Substituting 𝑦=𝑒𝑟𝑡y=e rt into the differential equation gives:42𝑟3𝑒𝑟𝑡+13𝑟2𝑒𝑟𝑡+𝑟𝑒𝑟𝑡=042r 3 e rt +13r 2 e rt +re rt =0Factor out 𝑒𝑟𝑡e rt :𝑒𝑟𝑡(42𝑟3+13𝑟2+𝑟)=0e rt (42r 3 +13r 2 +r)=0Since 𝑒𝑟𝑡≠0e rt =0, we have:42𝑟3+13𝑟2+𝑟=042r 3 +13r 2 +r=0Factor out 𝑟r:𝑟(42𝑟2+13𝑟+1)=0r(42r 2 +13r+1)=0This gives us one root 𝑟=0r=0. To find the other roots, solve the quadratic equation:42𝑟2+13𝑟+1=042r 2 +13r+1=0Use the quadratic formula 𝑟=−𝑏±𝑏2−4𝑎𝑐2𝑎r= 2a−b± b 2 −4ac​ ​ :𝑟=−13±132−4⋅42⋅12⋅42r= 2⋅42−13± 13 2 −4⋅42⋅1​ ​ 𝑟=−13±169−16884r= 84−13± 169−168​ ​ 𝑟=−13±184r= 84−13±1​ This gives us two roots:𝑟=−1284=−17r= 84−12​ =− 71​ 𝑟=−1484=−214=−16r= 84−14​ =− 142​ =− 61​ So, the roots are 𝑟=0r=0, 𝑟=−17r=− 71​ , and 𝑟=−16r=− 61​ . 2. **Form the general solution:** The general solution to the differential equation is:𝑦(𝑡)=𝐶1𝑒0𝑡+𝐶2𝑒−17𝑡+𝐶3𝑒−16𝑡y(t)=C 1​ e 0t +C 2​ e − 71​ t +C 3​ e − 61​ t 𝑦(𝑡)=𝐶1+𝐶2𝑒−17𝑡+𝐶3𝑒−16𝑡y(t)=C 1​ +C 2​ e − 71​ t +C 3​ e − 61​ t 3. **Apply the initial conditions:** Use the initial conditions to find 𝐶1C 1​ , 𝐶2C 2​ , and 𝐶3C 3​ . - 𝑦(0)=−7y(0)=−7:𝐶1+𝐶2+𝐶3=−7C 1​ +C 2​ +C 3​ =−7- 𝑦′(0)=2y ′ (0)=2:𝑦′(𝑡)=−17𝐶2𝑒−17𝑡−16𝐶3𝑒−16𝑡y ′ (t)=− 71​ C 2​ e − 71​ t − 61​ C 3​ e − 61​ t 𝑦′(0)=−17𝐶2−16𝐶3=2y ′ (0)=− 71​ C 2​ − 61​ C 3​ =2- 𝑦′′(0)=0y ′′ (0)=0:𝑦′′(𝑡)=(149𝐶2𝑒−17𝑡+136𝐶3𝑒−16𝑡)y ′′ (t)=( 491​ C 2​ e − 71​ t + 361​ C 3​ e − 61​ t )𝑦′′(0)=149𝐶2+136𝐶3=0y ′′ (0)= 491​ C 2​ + 361​ C 3​ =0We now have a system of equations:𝐶1+𝐶2+𝐶3=−7C 1​ +C 2​ +C 3​ =−7−17𝐶2−16𝐶3=2− 71​ C 2​ − 61​ C 3​ =2149𝐶2+136𝐶3=0491​ C 2​ + 361​ C 3​ =0

Suppose y(x)=7e−2x is a solution of the initial value problem dydx =−ky, y(0)=y0. What are the constants k and y0?k= help (numbers)y0=

t) Suppose y(x)=7e−2x is a solution of the initial value problem dydx =−ky, y(0)=y0. What are the constants k and y0?