Question

match the differential equation with its direction field (labeled i-iv). give reasons for your answer.

1. $y = y - 1$
2. $y = y^2 - x^2$
3. $y = y - x$
4. $y = y^3 - x^3$

i ii iii iv

direction field for the differential equation $y = 1 - y$ is shown in the accompanying figure. in each part, sketch the graph of the solution that satisfies the initial condition.
(a) $y(0) = -1$ (b) $y(0) = 1$ (c) $y(1) = 2$

a direction field for the differential equation $y = 2y - x$ is shown in the accompanying figure. in each part, sketch the graph of the solution that satisfies the initial condition.
(a) $y(1) = 1$ (b) $y(0) = -1$ (c) $y(-1) = 0$

Explanation:

Part 1: Match Differential Equations to Direction Fields

Step1: Analyze $y'=y-1$

Set $y'=0$: $y-1=0 \implies y=1$. This is a horizontal line where slopes are 0. Direction Field III has horizontal segments along $y=1$, so this matches III.

Step2: Analyze $y'=y^2-x^2$

Factor: $y'=(y-x)(y+x)$. Slopes are 0 when $y=x$ or $y=-x$. Direction Field I has 0 slopes along these two lines, so this matches I.

Step3: Analyze $y'=y-x$

Set $y'=0$: $y-x=0 \implies y=x$. This is a line where slopes are 0. Direction Field II has horizontal segments along $y=x$, so this matches II.

Step4: Analyze $y'=y^3-x^3$

Factor: $y'=(y-x)(y^2+xy+x^2)$. Slopes are 0 only when $y=x$ (since $y^2+xy+x^2>0$ for all real $x,y$ not both 0). Direction Field IV has 0 slopes only along $y=x$, so this matches IV.

Part 2: Sketch Solutions for $y'=1-y$

• (a) Start at $(0,-1)$, follow direction field: the solution is an increasing curve approaching $y=1$ as $x\to\infty$, decreasing as $x\to-\infty$.
• (b) Start at $(0,1)$: this is the equilibrium solution, so the graph is the horizontal line $y=1$.
• (c) Start at $(1,2)$, follow direction field: the solution is a decreasing curve approaching $y=1$ as $x\to\infty$, increasing as $x\to-\infty$.

Part 3: Sketch Solutions for $y'=2y-x$

• (a) Start at $(1,1)$, follow direction field: the solution is an increasing curve that grows as $x\to\infty$ and decreases as $x\to-\infty$.
• (b) Start at $(0,-1)$, follow direction field: the solution is a curve that decreases as $x\to-\infty$ and grows as $x\to\infty$, approaching the line $y=\frac{x}{2}+\frac{1}{4}$ (the equilibrium solution) asymptotically.
• (c) Start at $(-1,0)$, follow direction field: the solution is an increasing curve that grows as $x\to\infty$ and decreases as $x\to-\infty$.

Answer:

1. Matching:

• $y'=y-1$ $

ightarrow$ III

• $y'=y^2-x^2$ $

ightarrow$ I

• $y'=y-x$ $

ightarrow$ II

• $y'=y^3-x^3$ $

ightarrow$ IV

1. For $y'=1-y$:

• (a) Curve starting at $(0,-1)$, approaching $y=1$
• (b) Horizontal line $y=1$
• (c) Curve starting at $(1,2)$, approaching $y=1$

1. For $y'=2y-x$:

• (a) Curve starting at $(1,1)$, growing as $x\to\infty$
• (b) Curve starting at $(0,-1)$, approaching $y=\frac{x}{2}+\frac{1}{4}$
• (c) Curve starting at $(-1,0)$, growing as $x\to\infty$