Answer to Question #211671 in Differential Equations for Bilal Ur Rehman

Question

In the following problem use a suitable substitution to reduce the order of the given DE and then solve it

1. y''=2y'
2. y''+e^yy'2=0
3. y"+(1-1/y)y'²=0

Answer 1

1. y''=2y'

Substitution: $y' = p(y) \Rightarrow y'' = p\frac{{dp}}{{dy}}$

Then

$p\frac{{dp}}{{dy}} = 2p \Rightarrow \frac{{dp}}{{dy}} = 2 \Rightarrow dp = 2dy \Rightarrow p = 2y + 2{C_1} \Rightarrow \frac{{dy}}{{dx}} = 2y + 2{C_1} \Rightarrow \frac{{dy}}{{y + {C_1}}} = 2dx \Rightarrow \ln |y + {C_1}| = 2x - 2{C_2} \Rightarrow 2x = \ln |y + {C_1}| + 2{C_2} \Rightarrow x = \frac{1}{2}\ln |y + {C_1}| + {C_2}$

Answer: $x = \frac{1}{2}\ln |y + {C_1}| + {C_2}$

2.. y''+e^yy'2=0

Substitution: $y' = p(y) \Rightarrow y'' = p\frac{{dp}}{{dy}}$

Then

$p\frac{{dp}}{{dy}} + {e^y}{p^2} = 0 \Rightarrow p\frac{{dp}}{{dy}} = - {e^y}{p^2} \Rightarrow \frac{{dp}}{{dy}} = - {e^y}p \Rightarrow \frac{{dp}}{p} = - {e^y}dy \Rightarrow \ln p = - {e^y} + C \Rightarrow p = {e^{ - {e^y} + C}} \Rightarrow y' = {C_1}{e^{ - {e^y}}},\,\,{C_1} = {e^C} \Rightarrow \frac{{dy}}{{dx}} = {C_1}{e^{ - {e^y}}} \Rightarrow \frac{1}{{{C_1}}}{e^{{e^y}}}dy = dx \Rightarrow x = \frac{1}{{{C_1}}}\int {{e^{{e^y}}}dy} + {C_2} \Rightarrow x = \frac{1}{{{C_1}}}Ei({e^y}) + {C_2}$

Answer: $x = \frac{1}{{{C_1}}}Ei({e^y}) + {C_2}$

3.. y"+(1-1/y)y'²=0

Substitution: $y' = p(y) \Rightarrow y'' = p\frac{{dp}}{{dy}}$

Then

$p\frac{{dp}}{{dy}} + \left( {1 - \frac{1}{y}} \right){p^2} = 0 \Rightarrow p\frac{{dp}}{{dy}} = - \left( {1 - \frac{1}{y}} \right){p^2} \Rightarrow \frac{{dp}}{{dy}} = \left( {\frac{1}{y} - 1} \right)p \Rightarrow \frac{{dp}}{p} = \left( {\frac{1}{y} - 1} \right)dy \Rightarrow \ln p = \ln y - y + C \Rightarrow p = {e^{\ln y - y + C}} \Rightarrow p = {C_1}{e^{\ln y}}{e^{ - y}},\,\,{C_1} = {e^C} \Rightarrow y' = {C_1}y{e^{ - y}} \Rightarrow \frac{{dy}}{{dx}} = {C_1}y{e^{ - y}} \Rightarrow \frac{1}{{{C_1}}}\frac{{{e^y}dy}}{y} = dx \Rightarrow x = \frac{1}{{{C_1}}}\int {\frac{{{e^y}dy}}{y}} + {C_2} \Rightarrow x = \frac{1}{{{C_1}}}Ei(y) + {C_2}$

Answer: $x = \frac{1}{{{C_1}}}Ei(y) + {C_2}$