Solved: Consider the differential equation $\frac{dy}{dx} = g\left(\frac{y}{x}\right),... [IB Mathematics Analysis and Approaches (AA)]

Question

HLPaper 2

Consider the differential equation

$\frac{dy}{dx} = g\left(\frac{y}{x}\right), \quad x > 0$

The curve $y = g(x)$ for $x > 0$ has a gradient function given by

$\frac{dy}{dx} = \frac{y^2 + 3xy + 2x^2}{x^2}$

The curve passes through the point $(1, -1)$ .

Use the substitution $y=wx$ to show that

$\int \frac{dw}{g(w)-w}=\ln x+D$

where $D$ is an arbitrary constant.

[3]

Verified

Solution

Substitute $y = wx$ into the original equation:

$\frac{dy}{dx} = w + x\frac{dw}{dx} = g(w)$ M1
Rearrange to isolate $x\frac{dw}{dx}$ :

$x\frac{dw}{dx} = g(w) - w$ A1
Separate variables and integrate:

$\int \frac{dw}{g(w) - w} = \int \frac{dx}{x}$ A1
Integrate the right-hand side:

$\int \frac{dw}{g(w) - w} = \ln x + D$

Where $D$ is an arbitrary constant.

3 marks total

By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation

$y = x \tan(\ln(x)) - 1$ .

[6]

Verified

Solution

Method #1

Attempt to find $g(w)$ M1
$g(w) = w^2 + 3w + 2$ A1
Substitute $g(w)$ into $\int \frac{dw}{g(w)-w}$ M1
$\int \frac{dw}{g(w)-w} = \int \frac{dw}{w^2 + 2w + 2}$
Attempt to complete the square M1
$\int \frac{dw}{w^2 + 2w + 2} = \int \frac{dw}{(w+1)^2 + 1}$ A1
$\arctan (w+1) = \ln x+D$ A1

Method #2

Attempt to find $g(w)$ M1
$w + x\frac{dw}{dx} = w^2 + 3w + 2$ A1
$\int \frac{dw}{w^2 + 2w + 2} = \frac{dx}{x}$ M1
Attempt to complete the square M1
$\int \frac{dw}{w^2 + 2w + 2} = \int \frac{dx}{x}$ A1
$\arctan(w+1) = \ln x+D$ A1

Continuation

When $x=1$ , $w = -1$ (or $y=-1$ ) and so $D=0$ M1
Substitute for $w$ into their expression M1
$\arctan \left(\frac{y}{x}+1\right) = \ln x$
$\frac{y}{x}+1 = \tan \left(\ln x\right)$ A1
So $y = x\left(\tan \left(\ln x\right)-1\right)$

6 marks total

Use the differential equation $\frac{dy}{dx} = \frac{y^2 + 3xy + 2x^2}{x^2}$ to show that the points of zero gradient on the curve lie on two straight lines of the form $y = mx$ where the values of $m$ are to be determined.

[4]

Verified

Solution

Set $\frac{dy}{dx} = 0$ to find stationary points: $y^2 + 3xy + 2x^2 = 0$ M1
Attempt to solve $y^2 + 3xy + 2x^2 = 0$ for $y$ M1
Factorize or use quadratic formula: $(y+2x)(y+x) = 0$ or $y = \frac{-3x \pm \sqrt{(3x)^2 - 4(2)(x^2)}}{2}$ A1
Obtain solutions: $y = -2x$ and $y = -x$ A1

Alternative method:

M1: State $\frac{dy}{dx} = 0$
M1: Substitute $y = mx$ into $\frac{dy}{dx} = 0$
A1: Obtain $(m+2)(m+1) = 0$
A1: Solve $m = -2, -1 \Rightarrow y = -2x$ and $y = -x$

4 marks total

The curve has a point of inflexion at $x_2, y_2$ where $e^{-\frac{\pi}{2}} < x_2 < e^{\frac{\pi}{2}}$ . Determine the coordinates of this point of inflexion.

[6]

Verified

Solution

Method #1

Correct graph of $y=g'(x)$ (for approximately $e^{-\frac{\pi}{2}}<x<e^{\frac{\pi}{2}}$ ) with a local minimum point below the $x$ -axis A2

Award M1A1 for $\frac{dy}{dx}= \tan(\ln x)+\sec^2(\ln x)-1$

Attempts to find the $x$ -coordinate of the local minimum point on the graph of $y=g'(x)$ M1

Correct graph of $y=g''(x)$ (for approximately $e^{-\frac{\pi}{2}}<x<e^{\frac{\pi}{2}}$ ) showing the location of the $x$ -intercept A2

Award M1A1 for $\frac{d^2y}{dx^2}=\frac{\sec^2(\ln x)}{x}+\frac{2\sec^2(\ln x)\tan(\ln x)}{x}$

Attempts to find the $x$ -intercept M1

THEN

$x=0.629 = e^{-\arctan(\frac{1}{2})}$ A1
Attempts to find $g(0.629)$
The coordinates are $(0.629,-0.943)$ , $(e^{-\arctan(\frac{1}{2})},- \frac{3}{2}e^{-\arctan(\frac{1}{2})})$ A1

Method #2

Attempts implicit differentiation on $\frac{dy}{dx}$ to find $\frac{d^2y}{dx^2}$ M1
$\frac{d^2y}{dx^2}=\frac{(2y+3x)(x\frac{dy}{dx}-y)}{x^3}$ (or equivalent)
$\frac{d^2y}{dx^2}=0 \Rightarrow y=-\frac{3x}{2}$ ( $\frac{dy}{dx}\neq\frac{y}{x}$ ) A1
Attempts to solve $-\frac{3x^2}{2}=x\tan(\ln x)-1$ for $x$ where $e^{-\frac{\pi}{2}}<x<e^{\frac{\pi}{2}}$ M1
$x=0.629 = e^{-\arctan(\frac{1}{2})}$ A1
Attempts to find $g(0.629)$
The coordinates are $(0.629,-0.943)$ , $(e^{-\arctan(\frac{1}{2})},- \frac{3}{2}e^{-\arctan(\frac{1}{2})})$ A1

6 marks total

Question

HLPaper 2

Consider the differential equation

$\frac{dy}{dx} = g\left(\frac{y}{x}\right), \quad x > 0$

The curve $y = g(x)$ for $x > 0$ has a gradient function given by

$\frac{dy}{dx} = \frac{y^2 + 3xy + 2x^2}{x^2}$

The curve passes through the point $(1, -1)$ .

Use the substitution $y=wx$ to show that

$\int \frac{dw}{g(w)-w}=\ln x+D$

where $D$ is an arbitrary constant.

[3]

Verified

Solution

Substitute $y = wx$ into the original equation:

$\frac{dy}{dx} = w + x\frac{dw}{dx} = g(w)$ M1
Rearrange to isolate $x\frac{dw}{dx}$ :

$x\frac{dw}{dx} = g(w) - w$ A1
Separate variables and integrate:

$\int \frac{dw}{g(w) - w} = \int \frac{dx}{x}$ A1
Integrate the right-hand side:

$\int \frac{dw}{g(w) - w} = \ln x + D$

Where $D$ is an arbitrary constant.

3 marks total

By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation

$y = x \tan(\ln(x)) - 1$ .

[6]

Verified

Solution

Method #1

Attempt to find $g(w)$ M1
$g(w) = w^2 + 3w + 2$ A1
Substitute $g(w)$ into $\int \frac{dw}{g(w)-w}$ M1
$\int \frac{dw}{g(w)-w} = \int \frac{dw}{w^2 + 2w + 2}$
Attempt to complete the square M1
$\int \frac{dw}{w^2 + 2w + 2} = \int \frac{dw}{(w+1)^2 + 1}$ A1
$\arctan (w+1) = \ln x+D$ A1

Method #2

Attempt to find $g(w)$ M1
$w + x\frac{dw}{dx} = w^2 + 3w + 2$ A1
$\int \frac{dw}{w^2 + 2w + 2} = \frac{dx}{x}$ M1
Attempt to complete the square M1
$\int \frac{dw}{w^2 + 2w + 2} = \int \frac{dx}{x}$ A1
$\arctan(w+1) = \ln x+D$ A1

Continuation

When $x=1$ , $w = -1$ (or $y=-1$ ) and so $D=0$ M1
Substitute for $w$ into their expression M1
$\arctan \left(\frac{y}{x}+1\right) = \ln x$
$\frac{y}{x}+1 = \tan \left(\ln x\right)$ A1
So $y = x\left(\tan \left(\ln x\right)-1\right)$

6 marks total

[4]

Verified

Solution

Set $\frac{dy}{dx} = 0$ to find stationary points: $y^2 + 3xy + 2x^2 = 0$ M1
Attempt to solve $y^2 + 3xy + 2x^2 = 0$ for $y$ M1
Factorize or use quadratic formula: $(y+2x)(y+x) = 0$ or $y = \frac{-3x \pm \sqrt{(3x)^2 - 4(2)(x^2)}}{2}$ A1
Obtain solutions: $y = -2x$ and $y = -x$ A1

Alternative method:

M1: State $\frac{dy}{dx} = 0$
M1: Substitute $y = mx$ into $\frac{dy}{dx} = 0$
A1: Obtain $(m+2)(m+1) = 0$
A1: Solve $m = -2, -1 \Rightarrow y = -2x$ and $y = -x$

4 marks total

The curve has a point of inflexion at $x_2, y_2$ where $e^{-\frac{\pi}{2}} < x_2 < e^{\frac{\pi}{2}}$ . Determine the coordinates of this point of inflexion.

[6]

Verified

Solution

Method #1

Correct graph of $y=g'(x)$ (for approximately $e^{-\frac{\pi}{2}}<x<e^{\frac{\pi}{2}}$ ) with a local minimum point below the $x$ -axis A2

Award M1A1 for $\frac{dy}{dx}= \tan(\ln x)+\sec^2(\ln x)-1$

Attempts to find the $x$ -coordinate of the local minimum point on the graph of $y=g'(x)$ M1

Correct graph of $y=g''(x)$ (for approximately $e^{-\frac{\pi}{2}}<x<e^{\frac{\pi}{2}}$ ) showing the location of the $x$ -intercept A2

Award M1A1 for $\frac{d^2y}{dx^2}=\frac{\sec^2(\ln x)}{x}+\frac{2\sec^2(\ln x)\tan(\ln x)}{x}$

Attempts to find the $x$ -intercept M1

THEN

$x=0.629 = e^{-\arctan(\frac{1}{2})}$ A1
Attempts to find $g(0.629)$
The coordinates are $(0.629,-0.943)$ , $(e^{-\arctan(\frac{1}{2})},- \frac{3}{2}e^{-\arctan(\frac{1}{2})})$ A1

Method #2

Attempts implicit differentiation on $\frac{dy}{dx}$ to find $\frac{d^2y}{dx^2}$ M1
$\frac{d^2y}{dx^2}=\frac{(2y+3x)(x\frac{dy}{dx}-y)}{x^3}$ (or equivalent)
$\frac{d^2y}{dx^2}=0 \Rightarrow y=-\frac{3x}{2}$ ( $\frac{dy}{dx}\neq\frac{y}{x}$ ) A1
Attempts to solve $-\frac{3x^2}{2}=x\tan(\ln x)-1$ for $x$ where $e^{-\frac{\pi}{2}}<x<e^{\frac{\pi}{2}}$ M1
$x=0.629 = e^{-\arctan(\frac{1}{2})}$ A1
Attempts to find $g(0.629)$
The coordinates are $(0.629,-0.943)$ , $(e^{-\arctan(\frac{1}{2})},- \frac{3}{2}e^{-\arctan(\frac{1}{2})})$ A1

6 marks total

Method #1

Method #2

Continuation

Method #1

Method #2

Related topics

Method #1

Method #2

Continuation

Method #1

Method #2

Related topics