IBDP Maths AHL 1.11 Partial fractions AA HL Paper 2- Exam Style Questions- New Syllabus

(a)(i)
Rearrange the differential equation to make \(\dfrac{dy}{dx}\) the subject.

\(\dfrac{dy}{dx}=\dfrac{x^2+xy-3y^2}{x^2+xy}\)

When \(x=1\), \(y=\dfrac{3}{2}\).

\(\dfrac{dy}{dx}=\dfrac{1+\frac32-3\left(\frac32\right)^2}{1+\frac32}\)

\(\dfrac{dy}{dx}=\dfrac{-\frac{17}{4}}{\frac52}=-\dfrac{17}{10}\)

✅ Answer: \(\dfrac{dy}{dx}=-1.7\)

(a)(ii)
There are two equal steps from \(x=1\) to \(x=1.4\), so the step size is:

\(h=\dfrac{1.4-1}{2}=0.2\)

Euler’s formula is \(y_{n+1}=y_n+h\,f(x_n,y_n)\).

First step:

\(x_0=1,\quad y_0=1.5,\quad f(x_0,y_0)=-1.7\)

\(y_1=1.5+0.2(-1.7)=1.16\)

Therefore, \((x_1,y_1)=(1.2,1.16)\).

Second step:

\(f(1.2,1.16)=\dfrac{1.2^2+(1.2)(1.16)-3(1.16)^2}{1.2^2+(1.2)(1.16)}\)

\(f(1.2,1.16)=-0.425423\ldots\)

\(y_2=1.16+0.2(-0.425423\ldots)\)

\(y_2=1.074915\ldots\)

✅ Answer: \(y(1.4)\approx1.07\)

(a)(iii)
During the first step, \(y\) decreases by \(0.34\), whereas during the second step it decreases by only approximately \(0.0851\).

The gradient changes from \(-1.7\) to approximately \(-0.425\), so the gradient is increasing and becoming less negative.

✅ Answer: The solution curve is concave up for \(1\leq x\leq1.4\).

(b)(i)
Differentiate both sides of

\(\left(x^2+xy\right)\dfrac{dy}{dx}=x^2+xy-3y^2\)

with respect to \(x\).

For the left-hand side, use the product rule:

\(\dfrac{d}{dx}\left[\left(x^2+xy\right)\dfrac{dy}{dx}\right]\)

\(=\left(2x+x\dfrac{dy}{dx}+y\right)\dfrac{dy}{dx}+\left(x^2+xy\right)\dfrac{d^2y}{dx^2}\)

Differentiate the right-hand side:

\(\dfrac{d}{dx}(x^2+xy-3y^2)=2x+x\dfrac{dy}{dx}+y-6y\dfrac{dy}{dx}\)

Therefore:

\(\left(x^2+xy\right)\dfrac{d^2y}{dx^2}+\left(2x+y\right)\dfrac{dy}{dx}+x\left(\dfrac{dy}{dx}\right)^2\)

\(=2x+y+(x-6y)\dfrac{dy}{dx}\)

Rearranging gives:

\(\left(x^2+xy\right)\dfrac{d^2y}{dx^2}=2x+y-x\left(\dfrac{dy}{dx}\right)^2-(x+7y)\dfrac{dy}{dx}\)

✅ Hence, the required result is shown.

(b)(ii)
When \(x=1\), \(y=\dfrac32\) and \(\dfrac{dy}{dx}=-\dfrac{17}{10}\).

Substitute these values into the result from part (b)(i).

\(\left(1+\dfrac32\right)\dfrac{d^2y}{dx^2}=2+\dfrac32-\left(-\dfrac{17}{10}\right)^2-\left(1+7\left(\dfrac32\right)\right)\left(-\dfrac{17}{10}\right)\)

\(\dfrac52\dfrac{d^2y}{dx^2}=20.16\)

\(\dfrac{d^2y}{dx^2}=8.064\)

✅ Answer: \(\dfrac{d^2y}{dx^2}=\dfrac{1008}{125}=8.064\)

(c)
Write the two fractions over the common denominator \(1-4v^2\).

\(\dfrac{A}{1-2v}+\dfrac{B}{1+2v}=\dfrac{A(1+2v)+B(1-2v)}{1-4v^2}\)

Therefore:

\(1+v=A(1+2v)+B(1-2v)\)

Comparing coefficients gives:

\(A+B=1\)

\(2A-2B=1\)

Solving these simultaneous equations:

✅ Answer: \(A=\dfrac34\) and \(B=\dfrac14\)

(d)
Since the differential equation is homogeneous, let:

\(v=\dfrac{y}{x}\), so \(y=vx\)

and

\(\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\)

Substitute these expressions into the differential equation.

\(\left(x^2+x(vx)\right)\left(v+x\dfrac{dv}{dx}\right)=x^2+x(vx)-3(vx)^2\)

Divide by \(x^2\).

\((1+v)\left(v+x\dfrac{dv}{dx}\right)=1+v-3v^2\)

Expanding and simplifying:

\(x(1+v)\dfrac{dv}{dx}=1-4v^2\)

Separate the variables.

\(\dfrac{1+v}{1-4v^2}\,dv=\dfrac{1}{x}\,dx\)

Using the partial fractions found in part (c):

\(\displaystyle\int\left(\dfrac{3}{4(1-2v)}+\dfrac{1}{4(1+2v)}\right)dv=\displaystyle\int\dfrac{1}{x}\,dx\)

\(-\dfrac38\ln|1-2v|+\dfrac18\ln|1+2v|=\ln x+C\)

Since \(y>\dfrac{x}{2}\), we have \(v>\dfrac12\). Therefore, \(|1-2v|=2v-1\).

\(-3\ln(2v-1)+\ln(1+2v)=8\ln x+C_1\)

When \(x=1\) and \(y=\dfrac32\), \(v=\dfrac32\).

\(-3\ln2+\ln4=C_1\)

\(C_1=-\ln2\)

Therefore:

\(-3\ln(2v-1)+\ln(1+2v)=8\ln x-\ln2\)

Rearranging:

\(8\ln x+3\ln(2v-1)=\ln(1+2v)+\ln2\)

Exponentiating both sides:

\(x^8(2v-1)^3=2(1+2v)\)

Substitute \(v=\dfrac{y}{x}\).

\(x^8\left(\dfrac{2y-x}{x}\right)^3=2\left(\dfrac{x+2y}{x}\right)\)

Multiplying through by \(x\) and simplifying:

✅ Hence, \(x^6(2y-x)^3=2(x+2y)\).