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

Find y'' by implicit differentiation.7x2 + y2 = 5

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

Test if the following ODEs are exact, and if so, solve: 7(i) y′ = yx

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: \[ y(t) = 7te^{2t} - 7e^{2t} + 8e^{t} \]

7x−5y=17𝑥−5𝑦=1; solve for x.