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=

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

Find the solution of the differential equation that satisfies the given initial condition.dydx = xy, y(0) = −1

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

Solve the system:  y = x2 – 7,  y =  x - 7Question 28Select one:a.x = 0b.(0, -1)c.(0, -7)d.(0, -7) and (1, -6)e.None of these