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Wednesday, June 16, 2021

solve y log y dx – x dy = 0

 

ncert cbse mathematics  class 12th differential equations

exercise 9.4 variable separable differential equations type

 

 7. solve y log y dx – x dy = 0


This is a variable separable differential equation


separating the variables [dividing each term with y.log y . x


[dx/x ] -  [ dy / (y.log y ) ] = 0


integrating using integration formula


and rearranging [ dy / (y.log y ) ] =[ (1/y)dy / logy ] and noting

that integral of {(f ') / f }form is ln(f)  


log x - log{logy}+logC = 0

here log stands for ln (natural logarithm)

 

using properties of log


log[ Cx / logy] = 0


Cx / log y = e^0


Cx / logy =1


log y = Cx


or y = e^(Cx)

 

8. solve (x^5) {dy/dx} = − (y^5 )


This is a variable separable differential equation


separating the variables

dy/(y^5 ) = -dx /(x^5 )

(y^(-5) )dy =- (x^(-5) )dx

 

integrating using integration formula


[(y^(-4)]/(-4) = -[(x^(-4)]/(-4) +C/(-4)


so that

[(x^(-4)]+[(y^(-4)] =C


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 ncert cbse 12th mathematics

exercise 9.4

differential equations 

variable separable type of differential equations


solve dy/dx = [1-cosx]/[1+cosx]

solution


2. solve (dy/dx) = sqrt[4-(y^2)]

solution

3. solve (dy/dx)+y=1

solution

4. { (sec x)^2} tan y dx + { (sec y)^2} tan x dy = 0

 solution

   


5.solve  ( e^x + e^( –x)  ) dy – ( e^x – e^(–x)  ) dx = 0

solution

6. solve (dy/dx)== (1 +( x^2) ) (1 + (y^2) )

solution


7. solve y log y dx – x dy = 0

solution

8. solve (x^5) {dy/dx} = − (y^5 )

solution







 










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