Homogeneous equation (y = vx substitution)

6 Use the substitution \(y = v x\) to find the solution of the differential equation $$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), where \(\mathrm { f } ( x )\) is a polynomial in \(x\). [10]
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6 Use the substitution \(y = v x\) to find the solution of the differential equation $$x \frac { d y } { d x } = y + \sqrt { 9 x ^ { 2 } + y ^ { 2 } }$$ for which \(y = 0\) when \(x = 1\). Give your answer in the form \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), where \(\mathrm { f } ( x )\) is a polynomial in \(x\). [10]
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3. (a) Show that the substitution \(y = v x\) transforms the differential equation $$\frac { d y } { d x } = \frac { 3 x - 4 y } { 4 x + 3 y }$$ into the differential equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \frac { 3 v ^ { 2 } + 8 v - 3 } { 3 v + 4 }$$ (b) By solving differential equation (II), find a general solution of differential equation (I). (5)
(c) Given that \(y = 7\) at \(x = 1\), show that the particular solution of differential equation (I) can be written as $$( 3 y - x ) ( y + 3 x ) = 200$$ (5)(Total 14 marks)

12. (a) Use the substitution \(y = v x\) to transform the equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + y ) ( x + y ) } { x ^ { 2 } } , x > 0$$ into the equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( 2 + v ) ^ { 2 }$$ (b) Solve the differential equation II to find \(\boldsymbol { v }\) as a function of \(\boldsymbol { x }\)
(c) Hence show that \(\quad y = - 2 x - \frac { x } { \ln x + c }\), where \(c\) is an arbitrary constant, is a general solution of the differential equation I.

1. (a) Show that the substitution \(y = v x\) transforms the differential equation

$$3 x y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 3 } + y ^ { 3 }$$ into the differential equation $$3 v ^ { 2 } x \frac { \mathrm {~d} v } { \mathrm {~d} x } = 1 - 2 v ^ { 3 }$$ (b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm { f } ( x )\). Given that \(y = 2\) at \(x = 1\),
(c) find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = 1\)

4 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } - y ^ { 2 } } { x y }$$

1. Use the substitution \(y = x z\), where \(z\) is a function of \(x\), to obtain the differential equation $$x \frac { \mathrm {~d} z } { \mathrm {~d} x } = \frac { 1 - 2 z ^ { 2 } } { z }$$
2. Hence show by integration that the general solution of the differential equation (A) may be expressed in the form \(x ^ { 2 } \left( x ^ { 2 } - 2 y ^ { 2 } \right) = k\), where \(k\) is a constant.

1 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ^ { 2 } + y ^ { 2 } } { x y } .$$

1. Use the substitution \(y = u x\), where \(u\) is a function of \(x\), to obtain the differential equation $$x \frac { \mathrm {~d} u } { \mathrm {~d} x } = \frac { 2 } { u } .$$
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

7. (a) Show that the substitution \(y = v x\) transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y } + \frac { 3 y } { x } , \quad x > 0 , \quad y > 0$$ into the differential equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = 2 v + \frac { 1 } { v } .$$ (b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm { f } ( x )\). Given that \(y = 3\) at \(x = 1\),
(c)find the particular solution of differential equation (I).(2)

4

1. Use the substitution \(y = x z\) to find the general solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y = x \cos \left( \frac { y } { x } \right)$$ giving your answer in a form without logarithms. (You may quote an appropriate result given in the List of Formulae (MF1).)
2. Find the solution of the differential equation for which \(y = \pi\) when \(x = 4\).

14. (a) Use the substitution \(y = v x\) to transform the equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + y ) ( x + y ) } { x ^ { 2 } } , x > 0$$ into the equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( 2 + v ) ^ { 2 }$$ (b) Solve the differential equation II to find \(v\) as a function of \(x\).
(c) Hence show that $$y = - 2 x - \frac { x } { \ln x + c } , \text { where } c \text { is an arbitrary constant, }$$ is a general solution of the differential equation I.
[0pt] [P4 January 2003 Qn 5]