ODE No. 78

2.78 ODE No. 78

$a \sin (α y (x) + β x) + b + y^{'} (x) = 0$ ✓ Mathematica : cpu = 0.921878 (sec), leaf count = 86

${{y (x) \to \frac{2 \tan^{- 1} (\frac{\sqrt{(β - α b)^{2} - a^{2} α^{2}} \tan (\frac{1}{2} (c_{1} - x) \sqrt{(β - α b)^{2} - a^{2} α^{2}}) - a α}{α b - β}) - β x}{α}}}$

✓ Maple : cpu = 2.266 (sec), leaf count = 89

${y (x) = \frac{1}{α} (- β x + 2 \arctan (\frac{- \tan (1 / 2 \sqrt{(- a^{2} + b^{2}) α^{2} - 2 α b β + β^{2}} (x -_C 1)) \sqrt{(- a^{2} + b^{2}) α^{2} - 2 α b β + β^{2}} - a α}{b α - β}))}$

Hand solution

$y^{'} = - a \sin (α y + β x) - b$

This is separable after transformation of $u = α y + β x$ , hence $u^{'} = α y^{'} + β$ or $y^{'} = \frac{1}{α} (u^{'} - β)$ . Therefore the above becomes $\begin{aligned} \frac{1}{α} (u^{'} - β) & = - a \sin (u) - b \\ u^{'} & = - α (a \sin (u) + b) + β \\ (1) & \frac{d u}{β - α (a \sin (u) + b)} & = d x \end{aligned}$

Using half angle tan transformation where $\tan (\frac{u}{2}) = t, \sin (u) = \frac{2 t}{t^{2} + 1}, d u = \frac{2}{1 + t^{2}} d t$ then $\begin{aligned} \int \frac{d u}{β - α (a \sin (u) + b)} & = \int \frac{2}{1 + t^{2}} \frac{d t}{β - α (a \frac{2 t}{t^{2} + 1} + b)} \\ = 2 \int \frac{d t}{β (t^{2} + 1) - α (a 2 t + b (t^{2} + 1))} \\ = 2 \int \frac{d t}{t β^{2} + β - (α a 2 t + t^{2} α b + α b)} \\ = 2 \int \frac{d t}{(β - α b) (\frac{t β^{2} - α a 2 t - t^{2} α b}{(β - α b)} + 1)} \\ = \frac{2}{(β - α b)} \int \frac{d t}{\frac{t β^{2} - α a 2 t - t^{2} α b}{(β - α b)} + 1} \\ = \frac{2}{(β - α b)} \frac{- (α b - β)}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}} \tanh^{- 1} (\frac{(t + \frac{a α}{b α - β}) (b α - β)}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}}) \\ = \frac{2}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}} \tanh^{- 1} (\frac{t (b α - β) + a α}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}}) \end{aligned}$

But $t = \tan (\frac{u}{2})$ therefore $\int \frac{d u}{β - α (a \sin (u) + b)} = \frac{2}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}} \tanh^{- 1} (\frac{\tan (\frac{u}{2}) (b α - β) + a α}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}})$ But $u = α y + β x$ , and the above becomes $\int \frac{d u}{β - α (a \sin (u) + b)} = \frac{2 \tanh^{- 1} (\frac{\tan (\frac{α y + β x}{2}) (b α - β) + a α}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}})}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}}$ Back to (1), therefore after integrating both sides $\frac{2 \tanh^{- 1} (\frac{\tan (\frac{α y + β x}{2}) (b α - β) + a α}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}})}{\sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}} = x + C$

Let $A = \sqrt{α^{2} a^{2} - (α^{2} b^{2} + β^{2} - 2 α b β)}$ Then

$\begin{aligned} \tanh^{- 1} (\frac{\tan (\frac{α y + β x}{2}) (b α - β) + a α}{A}) & = \frac{1}{2} A (x + C) \\ \frac{\tan (\frac{α y + β x}{2}) (b α - β) + a α}{A} & = \tanh (\frac{1}{2} A (x + C)) \\ \tan (\frac{α y + β x}{2}) (b α - β) + a α & = A \tanh (\frac{1}{2} A (x + C)) \\ \tan (\frac{α y + β x}{2}) & = \frac{A}{(b α - β)} \tanh (\frac{1}{2} A (x + C)) - \frac{a α}{(b α - β)} \\ \frac{α y + β x}{2} & = \arctan (\frac{A}{(b α - β)} \tanh (\frac{1}{2} A (x + C)) - \frac{a α}{(b α - β)}) \\ y & = \frac{2}{α} \arctan (\frac{A}{(b α - β)} \tanh (\frac{1}{2} A (x + C)) - \frac{a α}{(b α - β)}) - \frac{β x}{α} \end{aligned}$

Veriﬁcation

ode:=diff(y(x),x)=-a*sin(alpha*y(x)+beta*x)-b; 
A0:=sqrt(alpha^2*a^2-(alpha^2*b^2+beta^2-2*alpha*b*beta)); 
B0:=(alpha*b-beta); 
my_sol:=2/alpha*arctan(A0/B0*tanh((1/2)*A0*(x+_C1))-a*alpha/(B0))-beta*x/alpha; 
odetest(y(x)=my_sol,ode); 
0

[next] [prev] [prev-tail] [front] [up]