Use the substitution $y=vx$ to show that $x\frac{dv}{dx}=v^2-v-2$.

By solving the logistic differential equation, show that its solution can be expressed in the form

$kt=\ln \frac{P}{P_0}\left(\frac{N-P_0}{N-P}\right)$.

Solve the differential equation given that $y=-1$ when $x=1$. Give your answer in the form $y=f\left(x\right)$.

Solve the differential equation and show that a general solution is $x^2+y^2=cx$ where $c$ is a positive constant.

Show that $t$ + 1 is an integrating factor for this differential equation.

Given that $y\left(1\right)=1$, use Euler’s method with step length $h$ = 0.25 to find an approximation for $y\left(2\right)$. Give your answer to two significant figures.

Show that the general solution of this differential equation is $P=A{\text{e}}^{kt}$, where $A\in \mathbb{R}$.

$\underset{t\to \mathrm{\infty }}{\text{lim}}P$

the solution of the differential equation, giving your answer in the form $P=f\left(t\right)$.

Sketch the isoclines for $\frac{\text{d}y}{\text{d}x}=4$.

Find the value of $t$ at which the amount of salt in the tank is decreasing most rapidly.

Sketch the graph of $y=f\left(x\right)$ for $1⩽x⩽1.4$ .

Hence, or otherwise, solve the differential equation

$$\frac{\text{d}y}{\text{d}x}=\frac{x^2+3xy+y^2}{x^2},x>0,$$

given that $y=1$ when $x=1$. Give your answer in the form $y=g(x)$.

Consider the differential equation

$$\frac{\text{d}y}{\text{d}x}=f\left(\frac{y}{x}\right),x>0.$$

Use the substitution $y=vx$ to show that the general solution of this differential equation is

$$\int \frac{\text{d}v}{f(v)-v}=\ln x+\text{ Constant.}$$

Hence solve this differential equation. Give the answer in the form $y=f(x)$.

Given that the initial population is 1000, $L=10000$  and $m=0.003$, find the number of years it will take for the population to triple.

Show that $\frac{\text{d}P}{\text{d}t}=\frac{m}{L}P\left(L-P\right)$, where $m\in \mathbb{R}$.

the number of years it will take for the population to triple.

Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of $c$.

On the same set of axes, sketch and label isoclines for $\frac{\text{d}y}{\text{d}x}=-1, 0$ and $1$, and clearly indicate the value of each $y$-intercept.

Hence or otherwise, explain why the point $(1, 1)$ is a local minimum.

Find the solution of the differential equation $\frac{\text{d}y}{\text{d}x}=x-y$, which passes through the point $(1, 1)$. Give your answer in the form $y=f\left(x\right)$.

Sketch the graph of $y=f\left(x\right)$ on the same set of axes as part (a)(i).

Use Euler’s method, with step length $h=0.1$, to find an approximate value of $y$ when $x=1.4$.

Deduce a similar expression for $v(T+k)$ in terms of $k$.

Deduce the set of values for $p$ such that there are two points on the curve $y=f\left(x\right)$ where $\frac{\text{d}y}{\text{d}x}=0$. Give a reason for your answer.

the number of years it will take for the population to triple.

Show that $\sqrt{x^2+1}$ is an integrating factor for this differential equation.

Find the percentage error when $y\left(2\right)$ is approximated by the final rounded value found in part (a). Give your answer to two significant figures.

Solve the differential equation, for , giving your answer in the form $y=f\left(x\right)$.

the population after 10 years

By solving an appropriate differential equation, show that the particle’s velocity at time $t$ is given by $v\left(t\right)=(1+v_0){\text{e}}^{-t}-1$.

Show that the $x$-coordinate(s) of the points on the curve $y=f\left(x\right)$ where $\frac{\text{d}y}{\text{d}x}=0$ satisfy the equation $x^{p-1}=\frac{1}{p}$.

By using the result to part (b) (i), show that $v\left(T-k\right)={\text{e}}^k-1$.

Solve the differential equation giving your answer in the form $y=f(x)$.

With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture $f\left(1.4\right)$ will be less than, equal to, or greater than your answer in part (a).

Sketch the graph of $x$ versus $t$ for 0 ≤ $t$ ≤ 60 and hence find the maximum amount of salt in the tank and the value of $t$ at which this occurs.

Use the substitution $y=vx$ to show that $\int \frac{dv}{f\left(v\right)-v}=\ln x+C$ where $C$ is an arbitrary constant.

By solving an appropriate differential equation and using the result from part (b) (i), find an expression for $s_{\text{max}}$ in terms of $v_0$.

Solve the equation $\frac{\text{d}y}{\text{d}x}=1+\frac{y}{x}$ for $y\left(1\right)=1$.

Explain why the graph of $y=f\left(x\right)$ does not intersect the isocline $\frac{\text{d}y}{\text{d}x}=1$.

Find a prediction for the number of years it will take for the population of the world to double.

Hence, show that $v\left(T-k\right)+v\left(T+k\right)\ge 0$.

Use Euler’s method, with a step length of $0.1$, to find an approximate value of $y$ when $x=1.5$.

Solve the differential equation $\frac{dy}{dx}=\frac{\ln {\displaystyle }{\displaystyle 2}{\displaystyle x}}{x^2}-\frac{2y}{x}, x>0$, given that $y=4$ at $x=\frac{1}{2}$.

Give your answer in the form $y=f\left(x\right)$.

By substituting $Y=\frac{dy}{dt}$, show that $Y=B{\text{e}}^{2t}$ where $B$ is a constant.

Let the two values found in part (c)(ii) be ${\lambda }_1$ and ${\lambda }_2$.

Verify that $y=F{\text{e}}^{{\lambda }_{1}t}+G{\text{e}}^{{\lambda }_{2}t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

Hence, by solving this differential equation, show that $x\left(t\right)=\frac{200-40{\text{e}}^{-\frac{t}{4}}\left(t+5\right)}{t+1}$.

Find the equation of the isocline corresponding to $\frac{\text{d}y}{\text{d}x}=k$, where $k\ne 0$, $k\in \mathbb{R}$.

The rate of change of the amount of salt leaving the tank is equal to $\frac{x}{t+1}$.

Find the amount of salt that left the tank during the first 60 minutes.

Solve the differential equation, giving your answer in the form $f\left(x, y\right)=0$.

Show that there are no points of inflexion on the graph of $y$ against $x$.

Express $m^2-2m+4$ in the form ${\left(m-a\right)}^2+b$ , where $a\text{, }b\in \mathbb{Z}$.

Solve the differential equation $\frac{\text{d}P}{\text{d}t}=\frac{m}{L}P\left(L-P\right)$, giving your answer in the form $P=g\left(t\right)$.

Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.

Show that $1+x^2$ is an integrating factor for this differential equation.

By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation $y=x\left(\tan \left(\ln x\right)-1\right)$.

By solving the differential equation, show that $P=\frac{1200k}{\left(k-1200\right){\text{e}}^{-{\displaystyle \frac{t}{5}}}+1200}$.

Show that the time $T$ taken for the particle to reach $s_{\text{max}}$ satisfies the equation ${\text{e}}^T=1+v_0$.

The graph of $y$ against $x$ has a local maximum between $x=2$ and $x=3$. Determine the coordinates of this local maximum.

By solving the differential equation $\frac{dy}{dt}=y$, show that $y=A{\text{e}}^t$ where $A$ is a constant.

Show that $\frac{dx}{dt}-x=-A{\text{e}}^t$.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$.

By differentiating $\frac{dy}{dt}=-x+y$ with respect to $t$, show that $\frac{d^{2}y}{d{t}^2}=2\frac{{\displaystyle dy}}{{\displaystyle dt}}$.

Hence find $y$ as a function of $t$.

Hence show that $x=-\frac{B}{2}{\text{e}}^{2t}+C$, where $C$ is a constant.

Show that $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.

Find the two values for $\lambda$ that satisfy $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.