The solution to the ordinary differential equation y˙(t)=−2y(t),y(0)=1,𝑦˙(𝑡)=−2𝑦(𝑡),𝑦(0)=1, isy(t)=−2e−2t𝑦(𝑡)=−2𝑒−2𝑡y(t)=2e−2t𝑦(𝑡)=2𝑒−2𝑡y(t)=e−2t𝑦(𝑡)=𝑒−2𝑡y(t)=−e−2t

The solution to the ordinary differential equation y˙(t)+2y(t)=2,y(0)=0,𝑦˙(𝑡)+2𝑦(𝑡)=2,𝑦(0)=0, isy(t)=t𝑦(𝑡)=𝑡y(t)=1−e−2t𝑦(𝑡)=1−𝑒−2𝑡y(t)=2−2e−2t𝑦(𝑡)=2−2𝑒−2𝑡y(t)=1−2e−2t

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

The general solution to the differential equation 𝑑𝑦𝑑𝑥=𝑥𝑦 is 𝑦=±𝑥2+𝐶. Let 𝑦=𝑓(𝑥) be the particular solution to the differential equation with the initial condition 𝑓(−2)=−2. Which of the following is an expression for 𝑓(𝑥) and the domain for which the solution is valid?

Identify the type of the differential equation (as linear, homogeneous,exact, separable or Bernoulli) and solve the initial value problemdydt = 2 sin(t) − y, with y(π2)= 3

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