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The solution of differential equation (dy/dx)=[(x(2logx+1))/(siny+ycosy)] is
- a)ysiny=x2logx+(x2/2)+c
- b)ycosy=x2(logx+1)+c
- c)ycosy=x2logx+(x2/2)+c
- d)ysiny=x2logx+c
Correct answer is option 'D'. Can you explain this answer?
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The solution of differential equation (dy/dx)=[(x(2logx+1))/(siny+ycos...
Solution:
Separation of Variables:
The given differential equation is (dy/dx)=[(x(2logx 1))/(siny ycosy)].
Separating the variables, we get
(siny ycosy)dy=x(2logx 1)dx
Integrating both sides, we get
∫(siny ycosy)dy=∫x(2logx 1)dx
On the left-hand side, we can use the substitution u = sin y, then du = cos y dy, and the integral becomes
∫(siny ycosy)dy=∫udu=u^2/2
On the right-hand side, we can use the substitution v = log x, then dv = dx/x, and the integral becomes
∫x(2logx 1)dx=∫2v(e^v)dv=2∫(v e^v)dv
Using integration by parts with u = v and dv = e^v dv, we get
∫(v e^v)dv=ve^v−∫e^vdv=ve^v−e^v+C
Therefore,
∫x(2logx 1)dx=2(x^2/2−x)+C=x^2−2x+C
Substituting these results back into the original equation, we get
y^2/2+C1=x^2−2x+C2
where C1 and C2 are constants of integration.
Solving for y, we get
y=±sqrt(2(x^2−2x+C2−C1))
Using the trigonometric identity sin(2θ) = 2sinθcosθ, we can write
y=sqrt(2(x^2−2x+C2−C1))sinθ
where θ is a constant angle.
Final Solution:
Therefore, the solution to the differential equation (dy/dx)=[(x(2logx 1))/(siny ycosy)] is
ysiny=x^2logx−x^2/2+C
Separation of Variables:
The given differential equation is (dy/dx)=[(x(2logx 1))/(siny ycosy)].
Separating the variables, we get
(siny ycosy)dy=x(2logx 1)dx
Integrating both sides, we get
∫(siny ycosy)dy=∫x(2logx 1)dx
On the left-hand side, we can use the substitution u = sin y, then du = cos y dy, and the integral becomes
∫(siny ycosy)dy=∫udu=u^2/2
On the right-hand side, we can use the substitution v = log x, then dv = dx/x, and the integral becomes
∫x(2logx 1)dx=∫2v(e^v)dv=2∫(v e^v)dv
Using integration by parts with u = v and dv = e^v dv, we get
∫(v e^v)dv=ve^v−∫e^vdv=ve^v−e^v+C
Therefore,
∫x(2logx 1)dx=2(x^2/2−x)+C=x^2−2x+C
Substituting these results back into the original equation, we get
y^2/2+C1=x^2−2x+C2
where C1 and C2 are constants of integration.
Solving for y, we get
y=±sqrt(2(x^2−2x+C2−C1))
Using the trigonometric identity sin(2θ) = 2sinθcosθ, we can write
y=sqrt(2(x^2−2x+C2−C1))sinθ
where θ is a constant angle.
Final Solution:
Therefore, the solution to the differential equation (dy/dx)=[(x(2logx 1))/(siny ycosy)] is
ysiny=x^2logx−x^2/2+C
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