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 \(25y'' - 40y' + 16y = 0\) 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: \[y(t) = \left(2 + \frac{7}{5}t\right)e^{\frac{4}{5}t}\]

Solve the initial value problem \[ \begin{array}{ll} \frac{\mathrm{d}}{\mathrm{d} x} y(x)+4 y(x)=\mathrm{e}^{(-4 x)}(-3+4 x) ; & y(0)=3 \\ y(x)= & \text { 圆 } \end{array} \]

solve the initial value problem dy/dt = ((-2*y)/t) with y(3)=3

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: \[ \boxed{\frac{30}{7}}. \]

y” –3 y’ + 2y = 4   , y(0) = 1, y’(0) = 0