Differential equation given

3. $$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} x } - x y = 1$$

1. Show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = \frac { 1 } { 2 } \left( a \frac { \mathrm {~d} y } { \mathrm {~d} x } + b x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + c \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } \right)$$ where \(a , b\) and \(c\) are constants to be found. Given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = 2\)
2. find a series solution for \(y\) in ascending powers of ( \(x - 2\) ), up to and including the term in \(( x - 2 ) ^ { 4 }\). Write each term in its simplest form.
3. Use the solution to part (b) to find an approximate value for \(y\) when \(x = 2.1\), giving your answer to 3 decimal places.

6

1. Given that \(y = \cos ( \ln ( 1 + x ) )\), prove that
   1. \(\quad ( 1 + x ) \frac { \mathrm { d } y } { \mathrm {~d} x } = - \sin ( \ln ( 1 + x ) )\),
   2. \(( 1 + x ) ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 1 + x ) \frac { \mathrm { d } y } { \mathrm {~d} x } + y = 0\).
   3. Obtain an equation relating \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   4. Hence find the Maclaurin series for \(y\), up to and including the term in \(x ^ { 3 }\).

The displacement \(x\) metres of a particle at time \(t\) seconds is given by the differential equation $$\frac{d^2 x}{dt^2} + x + \cos x = 0.$$ When \(t = 0\), \(x = 0\) and \(\frac{dx}{dt} = \frac{1}{2}\). Find a Taylor series solution for \(x\) in ascending powers of \(t\), up to and including the term in \(t^4\). [5]

$$\frac{d^2 y}{dx^2} = e^x \left(2x \frac{dy}{dx} + y^2 + 1\right).$$

1. Show that $$\frac{d^4 y}{dx^4} = e^x \left[2x \frac{d^3 y}{dx^3} + 4 \frac{d^2 y}{dx^2} + 6y \frac{dy}{dx} + y^2 + 1\right],$$ where \(k\) is a constant to be found. [3]

Given that, at \(x = 0\), \(y = 1\) and \(\frac{dy}{dx} = 2\),

1. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x^4\). [4]

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

1. Show that $$\frac{d^2 y}{dx^2} + (1 - 2y) \frac{dy}{dx} = 3.$$ [2]

Given that \(y = 1\) at \(x = 1\),

1. find a series solution for \(y\) in ascending powers of \((x - 1)\), up to and including the term in \((x - 1)^3\). [8]

$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)

1. By differentiating equation (I) with respect to \(x\), show that

$$(x^2 + 1)\frac{d^2y}{dx^2} = 2y^2 + (1 - 2x)\frac{dy}{dx}$$ (I)

1. By differentiating equation (I) with respect to \(x\), show that $$(x^2 + 1)\frac{d^3y}{dx^3} = (1 - 4x)\frac{d^2y}{dx^2} + (4y - 2)\frac{dy}{dx}.$$ (3) Given that \(y = 1\) and \(\frac{dy}{dx} = 1\) at \(x = 0\),
2. find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x_3\).(4)
3. Use your series to estimate the value of \(y\) at \(x = -0.5\), giving your answer to two decimal places.(1)

Given that \(y = \cos\{\ln(1 + x)\}\), prove that

1. \((1 + x)\frac{\mathrm{d}y}{\mathrm{d}x} = -\sin\{\ln(1 + x)\}\), [1]
2. \((1 + x)^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (1 + x)\frac{\mathrm{d}y}{\mathrm{d}x} + y = 0\). [2]

Obtain an equation relating \(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\), \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}\). [2] Hence find Maclaurin's series for \(y\), up to and including the term in \(x^3\). [4] Verify that the same result is obtained if the standard series expansions for \(\ln(1 + x)\) and \(\cos x\) are used. [3]