**Example 1:** Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. y'=y-2x, (1,0).

**Remark:** Note that on y=2x, the direction is always horizontal. In the example, the intersection of the solution curve with y=2x is the maximum of the solution curve.

**Example 2:** **(a)** Use Euler’s method with the step h=0.1 to estimate the value of y(0.4), where y is the solution of the initial-value problem y'=y, y(0)=1. **(b)** We know that the exact solution of the initial-value problem in part (a) is y=e^x. Draw the graph of y=e^x, 0\leq x\leq 0.4 togther with the Euler approximation in part (a).

**Remark:** Because the second derivative is positive, Euler’s method underestimates the value of y(0.4).

**Example 3:** Solve the differential equation \frac{dy}{dt}=\frac{t}{ye^{y+t^2}}.

**Hint:** Separate the variables ye^ydy = te^{-t^2}dt.

**Problem 4:** Find the solution of the differential equation that satisfies the given initial condition. \frac{du}{dt} = \frac{2t+\sec^2t}{2u}, u(0)=-5.

**Hint:** Separate the variables 2udu = (2t+\sec^2t)dt.

**Problem 5:** Find an equation of the curve that passes through the point (0,1) and whose slope at (x,y) is xy.

**Hint:** Solve the initial-value problem dy/yx = xy, y(0)=1.

**Problem 6:** Solve the differential equation y'=x+y by making the change of variable u=x+y.

**Hint:** Note that du/dx = 1+y' = 1+u. Hence 1+x+y = 1+u = Ae^x.