Let's work through **Question 1** step-by-step: --- ### 1. Given \(\frac{dy}{dx} = (y-1)(y+1)^2(y+2)\): #### (a) **Sketch the phase line and classify all critical points.** **Critical points:** Set \((y-1)(y+1)^2(y+2) = 0\): - \(y-1=0 \implies y=1\) - \(y+1=0 \implies y=-1\) - \(y+2=0 \implies y=-2\) **Phase line analysis:** Check the sign of \(\frac{dy}{dx}\) in each interval: - **For \(y < -2\):** - \((y-1)<0\), \((y+1)<0\), \((y+2)<0\) - So: \((-)(-)^2(-) = -\) (since \((-)^2 = +\)), so the product is negative. - **Arrows point left**. - **For \(-2 < y < -1\):** - \((y-1)<0\), \((y+1)<0\), \((y+2)>0\) - \((-)(-)^2(+) = -(+)(+) = -\) - **Arrows point left**. - **For \(-1 < y < 1\):** - \((y-1)<0\), \((y+1)>0\), \((y+2)>0\) - \((-)(+)^2(+) = -(+)(+) = -\) - **Arrows point left**. - **For \(y > 1\):** - All terms positive. - \((+)(+)^2(+) = +\) - **Arrows point right**. **Classify equilibrium points:** - **At \(y = -2\):** - To left: arrows point left (towards \(-\infty\)) - To right: arrows point left (towards \(y = -2\)) - **Semi-stable (node)** - **At \(y = -1\):** - The factor is squared, so the flow does not change direction. - To left: arrows left - To right: arrows left - **Semi-stable (node)** - **At \(y = 1\):** - To left: arrows left - To right: arrows right - **Unstable (source/repeller)** --- #### (b) **Sketch the graph of solutions passing through the given points.** - For \(y(0) < 1\): Solution flows left, so \(y \to -\infty\) as \(x\) increases. - For \(y(0) > 1\): Solution grows towards \(+\infty\) as \(x\) increases. - For \(y(0) = 1\): Solution stays at \(y = 1\). --- #### (c) **Find \(\lim_{x \to \infty} y(x)\) for \(y(0) = -\frac{3}{2}\).** Since \(-2 < y(0) < -1\), by phase line, the solution will decrease without bound as \(x \to \infty\): \[ \boxed{\lim_{x \to \infty} y(x) = -\infty} \] --- #### (d) **State the solution to the IVP \(\frac{dy}{dx} = (y-1)(y+1)^2(y+2)\), \(y(0) = 1\).** Since \(y(0) = 1\) is an equilibrium point, the solution remains constant: \[ \boxed{y(x) = 1} \] --- ## **Summary of Answers** **(a) Critical points:** \(y = -2\) (semi-stable), \(y = -1\) (semi-stable), \(y = 1\) (unstable). **(b) Direction of flow:** Arrows point left for \(y < 1\), right for \(y > 1\). **(c) \(\lim_{x \to \infty} y(x) = -\infty\) for \(y(0) = -\frac{3}{2}\).** **(d) Solution for \(y(0) = 1\): \(y(x) = 1\).** --- Would you like to continue with Question 2?

Let's carefully work through **Question 1** and **Question 2**, providing detailed calculations and explanations for each part. --- ## **Question 1** Given the differential equation: \[ \frac{dy}{dx} = (y - 1)(y + 1)^2 (y + 2) \] --- ### **(a) Sketch the phase line and classify all equilibrium points** **Step 1: Find critical points** Set \(\frac{dy}{dx} = 0\): \[ (y - 1)(y + 1)^2 (y + 2) = 0 \] The roots are: - \(y = 1\) - \(y = -1\) - \(y = -2\) --- **Step 2: Determine the sign of \(\frac{dy}{dx}\) in each interval** - **Intervals:** 1. \(y < -2\) 2. \(-2 < y < -1\) 3. \(-1 < y < 1\) 4. \(y > 1\) - **Sign analysis:** | Interval | Sign of each factor | Overall sign | Flow direction | Arrow | |------------|---------------------|----------------|------------------|--------| | \(y < -2\) | \((y - 1)<0\), \((y+1)<0\), \((y+2)<0\) | \((-)(-)(-)= -\) | \(\frac{dy}{dx} < 0\) | Left (decreasing) | | \(-2 < y < -1\) | \((y - 1)<0\), \((y+1)<0\), \((y+2)>0\) | \((-)(-)(+)= -\) | \(\frac{dy}{dx} < 0\) | Left | | \(-1 < y < 1\) | \((y - 1)<0\), \((y+1)>0\), \((y+2)>0\) | \((-)(+)(+)= -\) | \(\frac{dy}{dx} < 0\) | Left | | \(y > 1\) | \((y - 1)>0\), \((y+1)>0\), \((y+2)>0\) | \((+)(+)(+)= +\) | \(\frac{dy}{dx} > 0\) | Right | --- ### **Step 3: Classify each equilibrium point** - **At \(y = -2\):** - To the left of \(-2\), \(\frac{dy}{dx}<0\) (flow down). - To the right of \(-2\), \(\frac{dy}{dx}<0\) (flow down). **Flow always towards \(-2\)** from both sides: **semi-stable node (attractor from one side, unstable from the other).** - **At \(y = -1\):** - The derivative is zero, and the sign does not change across \(-1\): - Just to the left of \(-1\), \(\frac{dy}{dx}<0\). - Just to the right, \(\frac{dy}{dx}<0\). **Flow always towards \(-1\)** from both sides: **semi-stable node.** - **At \(y = 1\):** - To the left of 1, \(\frac{dy}{dx}<0\). - To the right of 1, \(\frac{dy}{dx}>0\). **Flow away from \(y=1\)**: **unstable equilibrium (source).** --- ### **Phase line sketch:** - At \(y=-2\) and \(y=-1\): arrows point towards them from both sides (semi-stable). - At \(y=1\): arrows point away from it (unstable). --- ### **(b) Graph of solutions passing through given points** Given initial points: | \(x=0\) | \(y(0)\) | Behavior | |---------|----------|----------| | 3 | \(3\) | \(>1\) → solution increases, tends to \(+\infty\) as \(x \to \infty\). | | -2 | \(-2\) | At equilibrium, solution stays at \(-2\). | | -3 | \(-3\) | \(< -2\), solution decreases further towards \(-\infty\). | | -1 | \(-1\) | At equilibrium, solution stays at \(-1\). | | 3 | \(3\) | Same as first point, increases. | | 1 | \(1\) | At equilibrium, solution stays at 1. | | 2 | \(2\) | \(>1\), solution increases toward \(+\infty\). | **Summary:** - Solutions starting **less than \(-2\)** tend to \(-\infty\). - Solutions starting **between \(-2\) and \(-1\)** tend towards \(-1\) or \(-\infty\), depending on initial value. - Solutions starting **less than 1** tend to \(-\infty\) unless at \(-1\). - Solutions **at \(y=1\)** stay at 1. - Solutions **greater than 1** tend to \(+\infty\). --- ### **(c) Find \(\lim_{x \to \infty} y(x)\) for initial condition \(y(0) = -\frac{3}{2}\)** - \(y(0) = -1.5\), which lies **between \(-2\) and \(-1\)**. - From the phase line, solutions in this interval tend toward **\(-1\)** or \(-\infty\). **Check the stability:** - Since \(\frac{dy}{dx}<0\) in \(-2 < y < -1\), solutions decrease towards \(-\infty\). **Thus:** \[ \boxed{ \lim_{x \to \infty} y(x) = -\infty } \] --- ### **(d) Find the solution \(y(x)\) satisfying \(y(0)=1\)** - Because \(y=1\) is an equilibrium, the solution is: \[ \boxed{ y(x) = 1 } \] --- ## **Summary** | Part | Answer | |---------|---------| | (a) | Equilibria at \(y=-2\) (semi-stable), \(y=-1\) (semi-stable), \(y=1\) (unstable). | | (b) | Solutions flow towards \(-\infty\) for initial \(< -2\), towards \(-1\) for initial between \(-2\) and \(-1\), stay at \(1\) if started there, and grow unbounded for initial \(>1\). | | (c) | \(\boxed{\lim_{x \to \infty} y(x) = -\infty}\) for initial \(y(0) = -\frac{3}{2}\). | | (d) | \(\boxed{y(x) = 1}\) for initial \(y(0) = 1\). | --- ## **Question 2** ### **(a) Solve \(\frac{dy}{dx} = y(1 - y)\), with \(y(0)=2\)** --- ### **Step 1: Separation of variables** \[ \frac{dy}{y(1 - y)} = dx \] ### **Step 2: Partial fractions** \[ \frac{1}{y(1 - y)} = \frac{A}{y} + \frac{B}{1 - y} \] Multiply both sides by \(y(1 - y)\): \[ 1 = A(1 - y) + By \] Set: - For \(y=0\): \[ 1 = A(1-0) + B(0) \Rightarrow A=1 \] - For \(y=1\): \[ 1= A(0) + B(1) \Rightarrow B=1 \] Thus: \[ \frac{1}{y(1 - y)} = \frac{1}{y} + \frac{1}{1 - y} \] --- ### **Step 3: Integrate** \[ \int \left( \frac{1}{y} + \frac{1}{1 - y} \right) dy = \int dx \] \[ \int \frac{1}{y} dy + \int \frac{1}{1 - y} dy = x + C \] \[ \ln|y| - \ln|1 - y| = x + C \] or: \[ \ln \left| \frac{y}{1 - y} \right| = x + C \] ### **Step 4: Solve for \(y\)** \[ \frac{y}{1 - y} = K e^{x} \] where \(K = e^{C}\). --- ### **Step 5: Apply initial condition \(y(0)=2\)** \[ \frac{2}{1-2} = \frac{2}{-1} = -2 = K e^{0} = K \] So, \[ K = -2 \] ### **Final explicit solution:** \[ \frac{y}{1 - y} = -2 e^{x} \] or \[ y = (1 - y)(-2 e^{x}) \] \[ y = -2 e^{x} + 2 e^{x} y \] Bring all \(y\) terms to one side: \[ y - 2 e^{x} y = -2 e^{x} \] \[ y (1 - 2 e^{x}) = -2 e^{x} \] \[ \boxed{ y(x) = \frac{-2 e^{x}}{1 - 2 e^{x}} } \] --- ## **Summary for 2(a):** \[ \boxed{ \boxed{ y(x) = \frac{-2 e^{x}}{1 - 2 e^{x}} } } \] --- ### **(b) Solve \(\sqrt{y'} = 2(t + t y^*) e^{t}\), \(y(0) = 0\).** Note: The differential equation reads: \[ \sqrt{\frac{dy}{dt}} = 2(t + t y^*) e^{t} \] Assuming \(y^*\) is a typo, and the intended equation is: \[ \sqrt{\frac{dy}{dt}} = 2 t e^{t} \] **Step 1: Square both sides:** \[ \frac{dy}{dt} = \left( 2 t e^{t} \right)^2 = 4 t^2 e^{2t} \] --- ### **Step 2: Integrate** \[ dy = 4 t^2 e^{2t} dt \] Initial condition: \(y(0) = 0\). --- ### **Step 3: Integrate RHS** \[ y(t) = \int 4 t^2 e^{2t} dt \] Focus on: \[ I = \int t^2 e^{2t} dt \] Use **integration by parts twice**: - **First**, let \(u = t^2\), \(dv = e^{2t} dt\) \[ du = 2 t dt, \quad v = \frac{1}{2} e^{2t} \] - **Then:** \[ I = t^2 \cdot \frac{1}{2} e^{2t} - \int 2 t \cdot \frac{1}{2} e^{2t} dt = \frac{t^2}{2} e^{2t} - \int t e^{2t} dt \] Now, compute \(\int t e^{2t} dt\): - **Let** \(u = t\), \(dv= e^{2t} dt\) \[ du = dt, \quad v= \frac{1}{2} e^{2t} \] - **Then:** \[ \int t e^{2t} dt = t \cdot \frac{1}{2} e^{2t} - \int 1 \cdot \frac{1}{2} e^{2t} dt = \frac{t}{2} e^{2t} - \frac{1}{2} \int e^{2t} dt \] \[ = \frac{t}{2} e^{2t} - \frac{1}{2} \cdot \frac{1}{2} e^{2t} = \frac{t}{2} e^{2t} - \frac{1}{4} e^{2t} \] --- ### **Step 4: Final expression for \(I\):** \[ I = \frac{t^2}{2} e^{2t} - \left( \frac{t}{2} e^{2t} - \frac{1}{4} e^{2t} \right) = \frac{t^2}{2} e^{2t} - \frac{t}{2} e^{2t} + \frac{1}{4} e^{2t} \] \[ I = e^{2

Let's carefully work through **Question 1** and **Question 2**, providing detailed step-by-step calculations, explanations, and final answers. --- # Question 1 Given the differential equation: \[ \frac{dy}{dx} = (y - 1)(y + 1)^2 (y + 2) \] and initial conditions. --- ## Part (a): Sketch the phase line and classify all equilibrium points ### Step 1: Find all equilibrium points (critical points) Set \(\frac{dy}{dx} = 0\): \[ (y - 1)(y + 1)^2 (y + 2) = 0 \] Solutions: \[ \boxed{ y = -2,\quad y = -1,\quad y=1 } \] ### Step 2: Determine the sign of \(\frac{dy}{dx}\) in each interval Identify the sign of each factor in the intervals: - For \(y < -2\): - \((y-1)<0\), \((y+1)<0\), \((y+2)<0\) - \(\frac{dy}{dx} = (-)(-)^2(-) = -\) - **Flow: decreasing** (arrow points left) - For \(-2< y < -1\): - \((y-1)<0\), \((y+1)<0\), \((y+2)>0\) - Sign: \((-)(-)^2(+)= -\) - **Flow: decreasing** - For \(-1< y < 1\): - \((y-1)<0\), \((y+1)>0\), \((y+2)>0\) - Sign: \((-)(+)(+)= -\) - **Flow: decreasing** - For \(y > 1\): - \((y-1)>0\), \((y+1)>0\), \((y+2)>0\) - Sign: \((+)(+)(+)= +\) - **Flow: increasing** ### Step 3: Classify the equilibrium points - **At \(y=-2\):** - To the left: flow downward - To the right: flow downward - **Flow always toward \(-2\)** from both sides. - **Semi-stable (attracting from one side, repelling from the other).** - **At \(y=-1\):** - Sign does not change across \(-1\); flow is always downward. - **Semi-stable node.** - **At \(y=1\):** - To the left: flow downward - To the right: flow upward - **Unstable equilibrium (source).** ### **Phase line sketch summary:** - \(-2\): semi-stable (flow toward \(-2\)) - \(-1\): semi-stable - \(1\): unstable (flow away) --- ## Part (b): Graph of solutions passing through given points Initial points: - \(y(0) = 3\): starts above \(1\), solution increases to \(\infty\). - \(y(0) = -2\): solution stays constant. - \(y(0) = -3\): starts less than \(-2\), solution decreases toward \(-\infty\). - \(y(0) = -1\): steady state, solution stays at \(-1\). - \(y(0) = 1\): steady state, solution stays at \(1\). - \(y(0) = 2\): starts above 1, solution increases to \(\infty\). **Summary:** - Initial \(> 1\): solutions grow unbounded. - Initial \(-2\): solution is constant. - Initial less than \(-2\): solutions decrease toward \(-\infty\). - Initial at \(-1\): remains at \(-1\). - Initial at \(1\): remains at \(1\). --- ## Part (c): Find \(\lim_{x \to \infty} y(x)\) with \(y(0) = -\frac{3}{2}\) - \(-2 < y(0) = -1.5 < -1\). - From the phase line: - For \(-2< y < -1\), \(\frac{dy}{dx} < 0\), so \(y\) decreases toward \(-\infty\). - **Therefore:** \[ \boxed{ \lim_{x \to \infty} y(x) = -\infty } \] --- ## Part (d): Find the solution with initial condition \(y(0)=1\) - Since \(y=1\) is an equilibrium point, the solution remains constant: \[ \boxed{ y(x) = 1 } \] --- # **Summary for Question 1:** | Part | Answer | |---------|---------| | (a) | Equilibria at \(y=-2\) (semi-stable), \(y=-1\) (semi-stable), \(y=1\) (unstable). | | (b) | Solutions tend to \(-\infty\) for initial \(< -2\), tend to \(-1\) or \(\infty\) depending on initial position, stay at equilibria if started there. | | (c) | \(\lim_{x \to \infty} y(x) = -\infty\) for initial \(y(0)=-\frac{3}{2}\). | | (d) | \(y(x)=1\). | --- # Question 2 --- ## Part (a): Solve \(\frac{dy}{dx} = y(1 - y)\), with initial condition \(y(0)=2\) ### Step 1: Separate variables \[ \frac{dy}{y(1 - y)} = dx \] ### Step 2: Partial fractions \[ \frac{1}{y(1 - y)} = \frac{A}{y} + \frac{B}{1 - y} \] Multiply both sides by \(y(1 - y)\): \[ 1 = A(1 - y) + B y \] Set \(y=0\): \[ 1 = A(1) \Rightarrow A=1 \] Set \(y=1\): \[ 1= B(1) \Rightarrow B=1 \] Thus: \[ \frac{1}{y(1 - y)} = \frac{1}{y} + \frac{1}{1 - y} \] ### Step 3: Integrate both sides \[ \int \left(\frac{1}{y} + \frac{1}{1 - y}\right) dy = \int dx \] \[ \int \frac{1}{y} dy + \int \frac{1}{1 - y} dy = x + C \] \[ \ln|y| - \ln|1 - y| = x + C \] \[ \ln \left| \frac{y}{1 - y} \right| = x + C \] ### Step 4: Solve for \(y\) \[ \frac{y}{1 - y} = K e^{x} \] Use initial condition \(y(0)=2\): \[ \frac{2}{1 - 2} = \frac{2}{-1} = -2 = K e^{0} = K \] \[ K = -2 \] ### Final explicit solution: \[ \frac{y}{1 - y} = -2 e^{x} \] Solve for \(y\): \[ y = (1 - y)(-2 e^{x}) \] \[ y = -2 e^{x} + 2 e^{x} y \] Bring \(y\) to one side: \[ y - 2 e^{x} y = -2 e^{x} \] \[ y (1 - 2 e^{x}) = -2 e^{x} \] \[ \boxed{ \boxed{ y(x) = \frac{-2 e^{x}}{1 - 2 e^{x}} } } \] --- ## Part (b): Solve \(\sqrt{\frac{dy}{dt}} = 2(t + t y^*) e^{t}\), with initial condition \(y(0)=0\) **Note:** Assuming \(y^*\) is a typo or a constant; perhaps the equation is: \[ \sqrt{\frac{dy}{dt}} = 2 t e^{t} \] --- ### Step 1: Square both sides \[ \frac{dy}{dt} = [2 t e^{t}]^2 = 4 t^2 e^{2t} \] ### Step 2: Integrate \[ dy = 4 t^2 e^{2t} dt \] Initial condition: \(y(0)=0\). ### Step 3: Compute \(\int 4 t^2 e^{2t} dt\) Focus on \(I = \int t^2 e^{2t} dt\). Use integration by parts twice: - First, set \(u = t^2\), \(dv = e^{2t} dt\): \[ du = 2 t dt, \quad v= \frac{1}{2} e^{2t} \] \[ I = t^2 \cdot \frac{1}{2} e^{2t} - \int 2 t \cdot \frac{1}{2} e^{2t} dt = \frac{t^2}{2} e^{2t} - \int t e^{2t} dt \] - Next, evaluate \(\int t e^{2t} dt\): Set \(u = t\), \(dv= e^{2t} dt\): \[ du= dt, \quad v= \frac{1}{2} e^{2t} \] \[ \int t e^{2t} dt = t \cdot \frac{1}{2} e^{2t} - \int 1 \cdot \frac{1}{2} e^{2t} dt = \frac{t}{2} e^{2t} - \frac{1}{2} \int e^{2t} dt \] \[ = \frac{t}{2} e^{2t} - \frac{1}{2} \cdot \frac{1}{2} e^{2t} = \frac{t}{2} e^{2t} - \frac{1}{4} e^{2t} \] Putting back into \(I\): \[ I = \frac{t^2}{2} e^{2t} - \left(\frac{t}{2} e^{2t} - \frac{1}{4} e^{2t}\right) = \frac{t^2}{2} e^{2t} - \frac{t}{2} e^{2t} + \frac{1}{4} e^{2t} \] Factor out \(e^{2t}\): \[ I = e^{2t} \left(\frac{t^2}{2} - \frac{t}{2} + \frac{1}{4}\right) \] Multiply numerator and denominator by 4 to combine: \[ I = e^{2t} \cdot \frac{2 t^2 - 2 t + 1}{4} \] --- ### **Step 4: Final expression for \(y(t)\):** \[ y(t) = \int 4 t^2 e^{2t} dt = 4 \cdot I = 4 \times \left[ e^{2t} \frac{2 t^2 - 2 t + 1}{4} \right] = e^{2t} (2 t^2 - 2 t + 1) \] Apply initial condition \(y(0)=0\): \[ y(0) = e^{0} (0 - 0 + 1) = 1 \] But initial condition states \(y(0)=0\), so to match the initial condition, **we need an additive constant \(C\)**: \[ y(t) = e^{2t}(2 t^2 - 2 t + 1) + C \] At \(t=0\): \[ 0 = 1 \times (0 - 0 + 1) + C \Rightarrow 0=1 + C \Rightarrow C = -1 \] **Final solution:** \[ \boxed{ \boxed{ y(t) = e^{2t}(2 t^2 - 2 t + 1) - 1 } } \] --- # **Summary of Question 2:** ### (a) \[ \boxed{ y(x) = \frac{-2 e^{x}}{1 - 2 e^{x}} } \] ### (b) \[ \boxed{ y(t) = e^{2t}(2 t^2 - 2 t + 1) - 1 } \] --- # **Final notes:** - For **Question 1**, the phase line analysis helped classify equilibrium points and determine the long-term behavior of solutions. - For **Question 2**, separation of variables leads to explicit solutions, with careful integration and initial condition application. If you'd like further clarification or additional problems solved, feel free to ask!

Certainly! Here are all the final answers for both questions summarized clearly: --- ## **Question 1 Final Answers** - **Equilibrium points:** \(y = -2\) (semi-stable), \(y = -1\) (semi-stable), \(y=1\) (unstable). - **Phase line classification:** - For \(y < -2\): flow toward \(-\infty\). - For \(-2 < y < -1\): flow toward \(-1\). - For \(-1 < y < 1\): flow toward \(-1\). - For \(y > 1\): flow toward \(+\infty\). - **Behavior of solutions passing through given initial points:** - \(y(0) = 3\): solutions increase unboundedly as \(x \to \infty\). - \(y(0) = -2\): solution stays constant at \(-2\). - \(y(0) = -3\): solution decreases toward \(-\infty\). - \(y(0) = -1\): solution stays constant at \(-1\). - \(y(0) = 1\): solution stays constant at \(1\). - \(y(0) = 2\): solutions increase unboundedly as \(x \to \infty\). - **Limit as \(x \to \infty\) for initial \(y(0) = -\frac{3}{2}\):** \[ \boxed{\lim_{x \to \infty} y(x) = -\infty} \] - **Solution for initial condition \(y(0) = 1\):** \[ \boxed{y(x) = 1} \] --- ## **Question 2 Final Answers** - **(a)** Solution to \(\frac{dy}{dx} = y(1 - y)\), \(y(0)=2\): \[ \boxed{ \boxed{ y(x) = \frac{-2 e^{x}}{1 - 2 e^{x}} } } \] - **(b)** Solution to \(\sqrt{\frac{dy}{dt}} = 2 t e^{t}\), \(y(0)=0\): \[ \boxed{ \boxed{ y(t) = e^{2t}(2 t^2 - 2 t + 1) - 1 } } \] These are the complete final answers for both questions.