4.10d Second order homogeneous: auxiliary equation method

Show that the substitution \(y = \frac { 1 } { w }\) reduces the differential equation $$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 y \frac { \mathrm {~d} y } { \mathrm {~d} x } - 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 5 y ^ { 2 } = \left( 5 x ^ { 2 } + 4 x + 2 \right) y ^ { 3 }$$ to $$\frac { \mathrm { d } ^ { 2 } w } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} w } { \mathrm {~d} x } + 5 w = - 5 x ^ { 2 } - 4 x - 2$$ Find the general solution for \(w\) in terms of \(x\). Find a function f such that \(\lim _ { x \rightarrow \infty } \left( \frac { y } { \mathrm { f } ( x ) } \right) = 1\).

Given that $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x ( 1 + x ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 \left( 1 + 4 x + 2 x ^ { 2 } \right) y = 8 x ^ { 2 }$$ and that \(x ^ { 2 } y = z\), show that $$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} z } { \mathrm {~d} x } + 4 z = 8 x ^ { 2 }$$ Find the general solution for \(y\) in terms of \(x\). Describe the behaviour of \(y\) as \(x \rightarrow \infty\).

3

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5$$
2. Hence express \(y\) in terms of \(x\), given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) when \(x = 0\).

8

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that:
   1. \(x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\quad x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }\).
2. Hence find the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 0$$

7

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$$
3. Find the general solution of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10\).
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 8\) when \(x = 1\).

3

1. A differential equation is given by $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x$$ Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x }$$ transforms this differential equation into $$\frac { \mathrm { d } u } { \mathrm {~d} x } + \frac { 2 } { x } u = 3$$
2. Find the general solution of $$\frac { \mathrm { d } u } { \mathrm {~d} x } + \frac { 2 } { x } u = 3$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

5 It is given that \(y\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \mathrm { e } ^ { - 2 x }$$

1. Find the value of the constant \(p\) for which \(y = p x \mathrm { e } ^ { - 2 x }\) is a particular integral of the given differential equation.
2. Solve the differential equation, expressing \(y\) in terms of \(x\), given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).

7 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 8 x ^ { 2 } + 9 \sin x$$ (8 marks)

8

1. Given that \(x = \mathrm { e } ^ { t }\) and that \(y\) is a function of \(x\), show that $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$$
2. Hence show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
3. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
4. Hence solve the differential equation \(x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x\), given that \(y = \frac { 3 } { 2 }\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 }\) when \(x = 1\).
   (5 marks)

3 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 10 y = 26 \mathrm { e } ^ { x }$$ given that \(y = 5\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 11\) when \(x = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).
(10 marks)

7 It is given that, for \(x \neq 0 , y\) satisfies the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y ( 3 x + 2 ) = 18 x$$

1. Show that the substitution \(u = x y\) transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} u } { \mathrm {~d} x } + 9 u = 18 x$$
2. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y ( 3 x + 2 ) = 18 x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
   (8 marks)

3 It is given that the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 0$$ is \(y = \mathrm { e } ^ { x } ( A x + B )\). Hence find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 6 \mathrm { e } ^ { x }$$

7

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ giving your answer in the form \(y = \mathrm { f } ( t )\).
2. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }$$ (5 marks)
3. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 6 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 10 y = \mathrm { e } ^ { 2 t }$$ (2 marks)
4. Hence write down the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 12 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 40 x ^ { 3 } y = 4 x ^ { 3 } \mathrm { e } ^ { 2 x ^ { 2 } }$$ (l mark)

1 It is given that \(y\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 8 x - 10 - 10 \cos 2 x$$

1. Show that \(y = 2 x + \sin 2 x\) is a particular integral of the given differential equation.
2. Find the general solution of the differential equation.
3. Hence express \(y\) in terms of \(x\), given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = 0\).

6

1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 10 \mathrm { e } ^ { - 2 x } - 9$$ (10 marks)
2. Hence express \(y\) in terms of \(x\), given that \(y = 7\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow 0\) as \(x \rightarrow \infty\).

5 It is given that \(y\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 8 \sin x + 4 \cos x$$

1. Find the value of the constant \(k\) for which \(y = k \sin x\) is a particular integral of the given differential equation.
2. Solve the differential equation, expressing \(y\) in terms of \(x\), given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 4\) when \(x = 0\).
   (8 marks)

8

1. Given that \(x = t ^ { 2 }\), where \(t \geqslant 0\), and that \(y\) is a function of \(x\), show that:
   1. \(2 \sqrt { x } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }\);
   2. \(\quad 4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }\).
2. Hence show that the substitution \(x = t ^ { 2 }\), where \(t \geqslant 0\), transforms the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } - 3 y = 0$$ (2 marks)
3. Hence find the general solution of the differential equation $$4 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 1 + 2 \sqrt { x } ) \frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = 0$$ giving your answer in the form \(y = \mathrm { g } ( x )\).

2

1. Find the value of the constant \(k\) for which \(k \sin 2 x\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + y = \sin 2 x$$
2. Hence find the general solution of this differential equation.

7

1. Given that \(x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0\) and \(y\) is a function of \(x\), show that:
   1. \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }\);
      (2 marks)
   2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }\).
      (3 marks)
2. Hence show that the substitution \(x = t ^ { \frac { 1 } { 2 } }\) transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t$$ (2 marks)
3. Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

2

1. Find the values of the constants \(p\) and \(q\) for which \(p + q x \mathrm { e } ^ { - 2 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = 4 - 9 \mathrm { e } ^ { - 2 x }$$
2. Hence find the general solution of this differential equation.
3. Hence express \(y\) in terms of \(x\), given that \(y = 4\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow 0\) as \(x \rightarrow \infty\).

7

1. Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$\text { into } \quad \begin{aligned} x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x ) \\ \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t \end{aligned}$$ (7 marks)
2. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$ (11 marks)
3. Write down the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$

3

1. Find the values of the constants \(a , b\) and \(c\) for which \(a + b x + c x \mathrm { e } ^ { - 3 x }\) is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 3 x - 8 \mathrm { e } ^ { - 3 x }$$
2. Hence find the general solution of this differential equation.
3. Hence express \(y\) in terms of \(x\), given that \(y = 1\) when \(x = 0\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow - 1\) as \(x \rightarrow \infty\).

7 A differential equation is given by $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x , \quad 0 < x < \pi$$

1. Show that the substitution $$y = u \sin x$$ where \(u\) is a function of \(x\), transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + u = \sin 2 x$$
2. Hence find the general solution of the differential equation $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
   (6 marks)

4 Solve the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 2 \mathrm { e } ^ { - x }$$ given that \(y \rightarrow 0\) as \(x \rightarrow \infty\) and that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 3\) when \(x = 0\).
[0pt] [10 marks] \includegraphics[max width=\textwidth, alt={}, center]{0eb3e96e-528c-4a99-b164-31cc865f0d68-08_46_145_829_159}

6

1. By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$ giving your answer in the form \(u = \mathrm { f } ( x )\).
   [0pt] [6 marks]
2. Show that the substitution \(u = x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }\) transforms the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ into $$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$
3. Hence, given that \(x > 0\), find the general solution of the differential equation $$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$ [2 marks]