2020_III_7_Pure

2020_III_7_Pure

(i)
Differentiating the second equation we have:

\displaystyle \frac {\mathrm d^2 y} {\mathrm dx^2} + g'(x)y + g(x) \frac {\mathrm dy} {\mathrm dx} = \frac {\mathrm du} {\mathrm dx}

Substituting the second equation into the first:

\displaystyle \frac {\mathrm du} {\mathrm dx} = h(x) - f(x) \frac {\mathrm dy} {\mathrm dx} - f(x)g(x)y

So:

\displaystyle \frac {\mathrm d^2 y} {\mathrm dx^2} + g'(x)y + g(x) \frac {\mathrm dy} {\mathrm dx} = h(x) - f(x) \frac {\mathrm dy} {\mathrm dx} - f(x)g(x)y

Rearranging gives:

\displaystyle \frac {\mathrm d^2 y} {\mathrm dx^2} + (g(x) + f(x)) \frac {\mathrm dy} {\mathrm dx} + (g'(x) + f(x)g(x))y = h(x)

as required.

(ii)

We have:

\displaystyle f(x) + g(x) = 1 + \frac 4 x

\displaystyle g'(x) + f(x)g(x) = \frac 2 x + \frac 2 {x^2}

Substituting the first equation in the second we have:

\displaystyle g'(x) + \left(1 + \frac 4 x - g(x)\right)g(x) = \frac 2 x + \frac 2 {x^2}

So that:

\displaystyle g'(x) + \left(1 + \frac 4 x\right)g(x) - (g(x))^2 = \frac 2 x + \frac 2 {x^2}

If g(x) = kx^n then g'(x) = nkx^{n - 1}, so:

\displaystyle nkx^{n - 1} + \left(1 + \frac 4 x\right)kx^n - k^2 x^{2 n} = \frac 2 x + \frac 2 {x^2}

By inspection, a possible value of n is -1, we can therefore write:

\displaystyle -\frac k {x^2} + \left(1 + \frac 4 x\right) \frac k x - \frac {k^2} {x^2} = -\frac {k} {x^2} + \frac k x + \frac {4k} {x^2} - \frac {k^2} {x^2} = \frac 2 x + \frac 2 {x^2}

By comparing coefficients, we see k = 2.

Since:

f(x) + g(x) = 1 + \dfrac 4 x

we have:

\displaystyle f(x) = 1 + \frac 4 x - \frac 2 x = 1 + \frac 2 x

Clearly h(x) = 4x + 12.

Giving the two differential equations:

\displaystyle \frac {\mathrm du} {\mathrm dx} + \left(1 + \frac 2 x\right)u = 4x + 12

and:

\displaystyle \frac {\mathrm dy} {\mathrm dx} + \frac 2 x y = u

We can solve the first by integrating factor, say:

\displaystyle e^{\int \left(1 + \frac 2 x\right) \mathrm dx} = e^{x + \ln(x^2)} = x^2 e^x

With this we rewrite the first differential equation:

\displaystyle x^2 e^x \frac {\mathrm du} {\mathrm dx} + \left(1 + \frac 2 x\right)x^2 e^x y = (4x + 12)x^2 e^x

That is:

\displaystyle x^2 e^x \frac {\mathrm du} {\mathrm dx} + (x^2 + 2 x)e^x u = (4x^3 + 12 x^2)e^x

The LHS is an exact derivative:

\displaystyle x^2 e^x \frac {\mathrm du} {\mathrm dx} + (x^2 + 2 x)e^x u = \frac {\mathrm d} {\mathrm dx} (x^2 e^x u)

as is the RHS, noticing that in a similar vein:

\displaystyle \frac {\mathrm d} {\mathrm dx} (x^3 e^x) = (x^3 + 3x^2)e^x

So the equation can be rewritten:

\displaystyle \frac {\mathrm d} {\mathrm dx} (x^2 e^x u) = 4 \frac {\mathrm d} {\mathrm dx} (x^3 e^x)

so:

x^2 e^x u = 4 x^3 e^x + C

that is:

\displaystyle u = 4 x + \frac C {x^2 e^x}

Substituting this into our second differential equation we have:

\displaystyle \frac {\mathrm dy} {\mathrm dx} + \frac 2 x y = 4 x + \frac C {x^2 e^x}

Using an integrating factor of:

\displaystyle e^{\int \frac 2 x \mathrm dx} = e^{\ln(x^2)} = x^2

we have:

\displaystyle x^2 \frac {\mathrm dy} {\mathrm dx} + 2 x y = 4 x^3 + Ce^{-x}

Noting again the LHS is an exact derivative:

\displaystyle x^2 \frac {\mathrm dy} {\mathrm dx} + 2xy = \frac {\mathrm d} {\mathrm dx} (x^2 y)

and:

\displaystyle \int \left(4x^3 + Ce^{-x}\right) \mathrm dx = x^4 - Ce^{-x} + D

so:

\displaystyle x^2y = x^4 - \frac C {e^x} + D

giving:

\displaystyle y = x^2 - \frac C {x^2 e^x} + \frac D {x^2}

We have:

\displaystyle y(1) = 1 - \frac C e + D = 5

We also have (from the second differential equation of (i)):

-3 + 5 g(1) = u(1)

That is:

\displaystyle -3 + 5 \times 2 = 4 + \frac C e

So:

C = 3e

Giving:

D = 5 - 1 + 3 = 7

Hence:

\displaystyle y = x^2 - \frac {3 e} {x^2 e^x} + \frac 7 {x^2}

1 Like

:earth_africa: :earth_americas: :earth_asia: This is a very nice one indeed! :earth_africa: :earth_americas: :earth_asia:

Marks breakdown on each part?

Is this something that would be helpful? I can think about how to do it at scale

I think Part A is 2-3 marks
then those simultaneous equations 1 mark
the first order DE 1 mark
gā€™(x) expression 1 mark
steps 2 marks
Value of n and k 1 mark
f(x) 1 mark
The two DEs 1 mark
The 2 integrating factors 1 mark each
steps and integrations 4 marks
Substitute initial values, find constants 3 marks

Are the mark schemes available for 2020 yet?