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 given initial value problem, we need to solve the first-order linear differential equation: \[ t y' + (t + 1)y = t \] with the initial condition: \[ y(\ln 8) = 1 \] ### Step 1: Rewrite the Differential EquationFirst, rewrite the differential equation in standard form: \[ y' + \left(1 + \frac{1}{t}\right)y = 1 \] ### Step 2: Find the Integrating FactorThe integrating factor \(\mu(t)\) is given by: \[ \mu(t) = e^{\int \left(1 + \frac{1}{t}\right) dt} \] Calculate the integral: \[ \int \left(1 + \frac{1}{t}\right) dt = \int 1 \, dt + \int \frac{1}{t} \, dt = t + \ln|t| \] So the integrating factor is: \[ \mu(t) = e^{t + \ln|t|} = e^t \cdot |t| = te^t \] ### Step 3: Multiply the Differential Equation by the Integrating FactorMultiply both sides of the differential equation by the integrating factor \(te^t\): \[ te^t y' + (t + 1) e^t y = t e^t \] ### Step 4: Simplify the Left SideNotice that the left side is the derivative of \(te^t y\): \[ \frac{d}{dt} \left(te^t y\right) = t e^t \] ### Step 5: Integrate Both SidesIntegrate both sides with respect to \(t\): \[ \int \frac{d}{dt} \left(te^t y\right) dt = \int t e^t \, dt \] The left side simplifies to: \[ te^t y \] For the right side, use integration by parts. Let \(u = t\) and \(dv = e^t dt\). Then \(du = dt\) and \(v = e^t\): \[ \int t e^t \, dt = t e^t - \int e^t \, dt = t e^t - e^t + C = e^t (t - 1) + C \] So we have: \[ te^t y = e^t (t - 1) + C \] ### Step 6: Solve for \(y\) Divide both sides by \(te^t\): \[ y = \frac{e^t (t - 1) + C}{te^t} = \frac{t - 1}{t} + \frac{C}{te^t} = 1 - \frac{1}{t} + \frac{C}{te^t} \] ### Step 7: Apply the Initial ConditionUse the initial condition \(y(\ln 8) = 1\): \[ 1 = 1 - \frac{1}{\ln 8} + \frac{C}{(\ln 8)e^{\ln 8}} \] Since \(e^{\ln 8} = 8\), we have: \[ 1 = 1 - \frac{1}{\ln 8} + \frac{C}{8 \ln 8} \] Simplify and solve for \(C\): \[ 0 = -\frac{1}{\ln 8} + \frac{C}{8 \ln 8} \] \[ \frac{C}{8 \ln 8} = \frac{1}{\ln 8} \] \[ C = 8 \] ### Step 8: Write the Final SolutionSubstitute \(C = 8\) back into the general solution: \[ y = 1 - \frac{1}{t} + \frac{8}{te^t} \] So the solution to the initial value problem is: \[ y = 1 - \frac{1}{t} + \frac{8}{te^t} \]

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?

Solve the initial-value problem.x2y' + 2xy = ln(x),    y(1) = 7

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

Solve the differential equation.y' = x + 7y