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The general solution of the differential equation (1+tany)(dx−dy)+2xdy=0 is
- a)x(siny+cosy)=siny+cey
- b)x(siny+cosy)=siny+ce−y
- c)y(sinx+cosx)=sinx+cex
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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The general solution of the differential equation(1+tany)(dx−dy)...
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The general solution of the differential equation(1+tany)(dx−dy)...
We have, (1+tany)(dx−dy)+2xdy=0
⇒(1+tany)dx=(1+tany−2x)dy
It is a linear differential equation.
=(cosy+siny)ey
So, the solution is
xey (sin y + cos y) = ∫ey (sin y + cos y)dy + c
⇒ xey(sin y+cos y)=eysin y + c
⇒ x(sin y + cos y) = sin y + ce−y.
⇒(1+tany)dx=(1+tany−2x)dy
It is a linear differential equation.
=(cosy+siny)ey
So, the solution is
xey (sin y + cos y) = ∫ey (sin y + cos y)dy + c
⇒ xey(sin y+cos y)=eysin y + c
⇒ x(sin y + cos y) = sin y + ce−y.
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The general solution of the differential equation(1+tany)(dx−dy)+2xdy=0 isa)x(siny+cosy)=siny+ceyb)x(siny+cosy)=siny+ce−yc)y(sinx+cosx)=sinx+cexd)None of theseCorrect answer is option 'B'. Can you explain this answer?
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