# Recitation 13

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 $1+x+y = 1+u = Ae^x$.