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: \[ y = 1 - \frac{1}{t} + \frac{8}{te^t} \]

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: \[ y = 1 - \frac{1}{t} + \frac{8}{te^t} \]

Y'=(y ln(y)+(yx^2))/-((x)+(2y^2)) ,y(3)=1 Find the solution of the following initial value problem.

Solve the initial-value problem.x2y' + 2xy = ln(x),    y(1) = 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: \[ y(t) = 7te^{2t} - 7e^{2t} + 8e^{t} \]

Solve the following initial value problem:𝑦=∫(3𝑥2−6)d𝑥,𝑦(1)=2