Explicit differential equation series solution

4. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } - x$$

1. Show that $$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = A y \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } + B \frac { \mathrm {~d} y } { \mathrm {~d} x } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$ where \(A\) and \(B\) are integers to be determined. Given that \(y = 1\) at \(x = - 1\)
2. determine the Taylor series solution for \(y\), in ascending powers of \(( x + 1 )\) up to and including the term in \(( x + 1 ) ^ { 4 }\), giving each coefficient in simplest form.

7. $$( 1 + 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x } = x + 4 y ^ { 2 }$$

1. Show that $$( 1 + 2 x ) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 1 + 2 ( 4 y - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$
2. Differentiate equation 1 with respect to \(x\) to obtain an equation involving $$\frac { \mathrm { d } ^ { 3 } } { \mathrm {~d} x ^ { 3 } } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x } , \quad x \text { and } y .$$ Given that \(y = \frac { 1 } { 2 }\) at \(x = 0\),
3. find a series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
   (6)(Total 11 marks)

3. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } - y ^ { 2 } , \quad y = 1 \text { at } x = 0 \text {. (I) }$$ (b) By differentiating (I) twice with respect to \(x\), show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } + 2 y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 2 = 0$$ (c) Hence, for (I), find the series solution for \(\boldsymbol { y }\) in ascending powers of \(\boldsymbol { x }\) up to and including the term in \(\boldsymbol { x } ^ { \mathbf { 3 } }\). (4)

7. $$\frac { \mathrm { d } ^ { 2 x } } { \mathrm {~d} t ^ { 2 } } + 3 \sin x = 0 . \quad \text { At } t = 0 , \quad x = 0 \quad \text { and } \quad \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4$$ (b) Find a series solution for \(x\), in ascending powers of \(t\), up to and including the term in \(t ^ { 3 }\).
(c) Use your answer to (b) to obtain an estimate of \(x\) at \(t = 0.3\).
(2)(Total 11 marks)

2. The displacement \(x\) metres of a particle at time \(t\) seconds is given by the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + x + \cos x = 0$$ When \(t = 0 , x = 0\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 2 }\).
Find a Taylor series solution for \(x\) in ascending powers of \(t\), up to and including the term in \(t ^ { 3 }\).

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

1. Show that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 1 - 2 y ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 3$$ Given that \(y = 1\) at \(x = 1\),
2. find a series solution for \(y\) in ascending powers of ( \(x - 1\) ), up to and including the term in \(( x - 1 ) ^ { 3 }\).

4. Use the Taylor Series method to find the series solution, ascending up to and including the term in \(x ^ { 3 }\), of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } = 3 x + 4$$ given that \(\frac { \mathrm { dy } } { \mathrm { dx } } = y = 1\) at \(x = 0\).
(Total 8 marks)

4. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y \frac { \mathrm {~d} y } { \mathrm {~d} x } = x , \quad y = 0 , \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 \text { at } x = 1$$ Find a series solution of the differential equation in ascending powers of ( \(x - 1\) ) up to and including the term in \(( x - 1 ) ^ { 3 }\).

11

1. At all points \(( x , y )\) on the curve \(C , \frac { \mathrm {~d} y } { \mathrm {~d} x } + x y = 0\).
   1. Prove by induction that, for all integers \(n \geqslant 1\), $$\frac { \mathrm { d } ^ { n + 1 } y } { \mathrm {~d} x ^ { n + 1 } } + x \frac { \mathrm {~d} ^ { n } y } { \mathrm {~d} x ^ { n } } + n \frac { \mathrm {~d} ^ { n - 1 } y } { \mathrm {~d} x ^ { n - 1 } } = 0$$ where \(\frac { \mathrm { d } ^ { 0 } y } { \mathrm {~d} x ^ { 0 } } = y\).
   2. Given that \(y = 1\) when \(x = 0\), determine the Maclaurin expansion of \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 6 }\).
   3. Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } + x y = 0\) given that \(y = 1\) when \(x = 0\).
   4. Given that \(Z \sim \mathrm {~N} ( 0,1 )\), use your answers to parts (i) and (ii) to find an approximation, to 4 decimal places, to the probability \(\mathrm { P } ( Z \leqslant 1 )\).
      [0pt] [Note that the probability density function of the standard normal distribution is \(\mathrm { f } ( z ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - \frac { 1 } { 2 } z ^ { 2 } }\).]

$$\frac{d^2 y}{dx^2} + 4y - \sin x = 0$$ Given that \(y = \frac{1}{2}\) and \(\frac{dy}{dx} = \frac{1}{8}\) at \(x = 0\), find a series expansion for \(y\) in terms of \(x\), up to and including the term in \(x^5\). [5]