Home / IBDP Maths AI: Topic : Topic: AHL 5.16: first order differential equations: IB style Questions HL Paper 1

IBDP Maths AI: Topic : Topic: AHL 5.16: first order differential equations: IB style Questions HL Paper 1

Question 16. [Maximum mark: 8]

A particle P moves in a straight line, such that its displacement x at time t (t ≥ 0) is defined by the differential equation x ̇ = x cost(e−sin t ).

 At time t = 0 , x = \(\frac{1}{c}\) .

a. By using Euler’s method with a step length of 0.1, find an approximate value

for x when t = 0.3 . [3]

b. By solving the differential equation, find the percentage error in your approximation

for x when t = 0.3 . [5]

▶️Answer/Explanation

(a)  \(x_{n}=x_{n-1}+h f(x_{n-1},t_{n-1}\)

\(h = 0.1\),\( f (x,t)= xcost (e^{-sint}\)

\(x_{n}=x_{n-1}+ 0.1x_{n-1} cost_{n-1}e^{-sint}\)

\(x (0.3) ≈ 0.477 (0.476548…)\)

(b) EITHER

\(\int \frac{dx}{x}= \int cost(e^{-sint})dt +c\)

\(ln x = e^{-sint}+ c\)

\(t= 0, x= \frac{1}{e}\Rightarrow c=0\)

\(x = e^{-sint}\)  

\(x(0.3)=0.475\)

THEN percentage error \(= |\frac{0.476548.. – 0.475140..}{0.475140..}|\times 100\)

\(= 0.296%(2.96192..)\)

Question

Solve the differential equation \(x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + 2y = \sqrt {1 + {x^2}} \)

given that \(y = 1\) when \(x = \sqrt 3 \) .

▶️Answer/Explanation

Markscheme

Rewrite the equation in the form

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{2}{x}y = \frac{{\sqrt {1 + {x^2}} }}{x}\)     M1A1

Integrating factor \( = {{\rm{e}}^{\int {\left( {\frac{2}{x}} \right)} {\rm{d}}x}}\)     M1

\( = {{\rm{e}}^{2\ln x}}\)     (A1)

\( = {x^2}\)     A1

The equation becomes

\(\frac{{\rm{d}}}{{{\rm{d}}y}}(y{x^2}) = x\sqrt {1 + {x^2}} \)     M1

\(y{x^2} = \frac{1}{3}{(1 + {x^2})^{\frac{3}{2}}} + C\)     A1

\(3 = \frac{1}{3} \times 8 + C \to C = \frac{1}{3}\)    M1A1

\(\left[ {{\text{giving }}y{x^2} = \frac{1}{3}\left( {{{\left( {1 + {x^2}} \right)}^{\frac{3}{2}}} + 1} \right)} \right]\)

[9 marks]

Question

Solve the following differential equation\[(x + 1)(x + 2)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y = x + 1\]giving your answer in the form \(y = f(x)\) .

▶️Answer/Explanation

Markscheme

Rewrite the equation in the form

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{y}{{(x + 1)(x + 2)}} = \frac{1}{{x + 2}}\)     M1

Integrating factor \( = \exp \left( {\int {\frac{{{\rm{d}}x}}{{(x + 1)(x + 2)}}} } \right)\)     A1

\( = \exp \left( {\int {\left( {\frac{1}{{x + 1}} – \frac{1}{{x + 2}}} \right){\rm{d}}x} } \right)\)     M1A1

\( = \exp \ln \left( {\frac{{x + 1}}{{x + 2}}} \right)\)     A1

\( = \frac{{x + 1}}{{x + 2}}\)     A1

Multiplying by the integrating factor,

\(\left( {\frac{{x + 1}}{{x + 2}}} \right)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{y}{{{{(x + 2)}^2}}} = \frac{{x + 1}}{{{{(x + 2)}^2}}}\)     M1

\( = \frac{{x + 2}}{{{{(x + 2)}^2}}} – \frac{1}{{{{(x + 2)}^2}}}\)     A1

Integrating,

\(\left( {\frac{{x + 1}}{{x + 2}}} \right)y = \ln (x + 2) + \frac{1}{{x + 2}} + C\)     A1A1

\(y = \left( {\frac{{x + 2}}{{x + 1}}} \right)\left\{ {\ln (x + 2) + \frac{1}{{x + 2}} + C} \right\}\)     A1

[11 marks]

Question

Given that \(\frac{{{\rm{d}}x}}{{{\rm{d}}y}} + 2y\tan x = \sin x\) , and \(y = 0\) when \(x = \frac{\pi }{3}\) , find the maximum value of y.

▶️Answer/Explanation

Markscheme

integrating factor \( = {{\rm{e}}^{\int {2\tan xdx} }}\)    M1

\( = {{\rm{e}}^{2\ln \sec x}}\)     A1 

\( = {\sec ^2}x\)     A1

it follows that

\(y{\sec ^2}x = \int {\sin x{{\sec }^2}x{\rm{d}}x} \)     M1

\( = \int {\sec x\tan x{\rm{d}}x} \)     (A1)

\( = \sec x + C\)     A1

substituting,

\(0 = 2 + C\) so \(C = – 2\)     M1A1

the solution is

\(y = \cos x – 2{\cos ^2}x\)    A1

EITHER

using a GDC

maximum value of \(y\) is \(0.125\)     A2

OR

\(y’ = – \sin x + 4\sin x\cos x = 0\)     M1

\( \Rightarrow \cos x = \frac{1}{4}\) (or \(\sin x = 0\) which leads to a minimum)

\( \Rightarrow y = \frac{1}{8}\)     A1

[11 marks]

Question

a.(i)      Find the range of values of \(n\) for which \(\int_1^\infty  {{x^n}{\rm{d}}x} \) exists.

(ii)     Write down the value of \(\int_1^\infty  {{x^n}{\rm{d}}x} \) in terms of \(n\) , when it does exist.[7]

 

b.Find the solution to the differential equation

\((\cos x – \sin x)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + (\cos x + \sin x)y = \cos x + \sin x\) ,

given that \(y = – 1\) when \(x = \frac{\pi }{2}\) .[8]

▶️Answer/Explanation

Markscheme

(i)     \(\int_1^b {{x^n}{\rm{d}}x}  = \left[ {\frac{{{x^{n + 1}}}}{{n + 1}}} \right]_1^b\) , \(n \ne – 1\)     M1

\( = \frac{{{b^{n + 1}}}}{{n + 1}} – \frac{1}{{n + 1}}\)     A1

\(\int_1^b {{x^n}{\rm{d}}x}  = \left[ {\ln x} \right]_1^b = \ln b\) when \(n = – 1\)     A1

if \(n + 1 > 0,\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} – \frac{1}{{n + 1}}} \right]\) does not exist since \({b^{n + 1}}\) increases without limit     R1

if \(n + 1 < 0,\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} – \frac{1}{{n + 1}}} \right]\) exists since \({b^{n + 1}} \to 0\) as \(b \to \infty \)     R1

if \(n =  – 1\) , \(\mathop {\lim }\limits_{b \to \infty } \left[ {\ln b} \right]\) does not exist since \({\ln b}\) increases without limit     R1

(so integral exists when \(n < – 1\) )

(ii)     \(\int_1^b {{x^n}{\rm{d}}x}  = \frac{1}{{n + 1}}\) , \((n < – 1)\)     A1 

[7 marks]

a.

\((\cos x – \sin x)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + (\cos x + \sin x)y = \cos x + \sin x\)

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{\cos x + \sin x}}{{\cos x – \sin x}}y = \frac{{\cos x + \sin x}}{{\cos x – \sin x}}\)     M1

IF \({{\rm{e}}^{\int {\frac{{\cos x + \sin x}}{{\cos x – \sin x}}} {\rm{d}}x}} = {{\rm{e}}^{ – \ln (\cos x – \sin x)}} = \frac{1}{{\cos x – \sin x}}\)     M1A1A1

\(\frac{y}{{\cos x – \sin x}} = \int {\frac{{\cos x + \sin x}}{{{{(\cos x – \sin x)}^2}}}} {\rm{d}}x\)     (M1)

\(\frac{1}{{\cos x – \sin x}} + k\)     A1

Note: Award the above A1 even if \(k\) is missing. 

\(y = 1 + k(\cos x – \sin x)\)

\(x = \frac{\pi }{2}\) , \(y =  – 1\)

\( – 1 = 1 + k( – 1)\)     M1

\(k = 2\)

\(y = 1 + 2(\cos x – \sin x)\)     A1

Note: It is acceptable to solve the equation using separation of variables.

[8 marks]

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