Q. 215.0( 1 Vote )

Find the particul

Answer :

Given: y = 0 when


To find: the particular solution of the differential equation


given:


This is of the of the standard form, i.e.,



Where P=cotx and Q=4x cosec x


And we know the integrating factor is


IF=e∫P dx


Substituting the value of P in the IF, we get


IF=e∫cot x dx


But we know integration of cot x is log(sinx), substituting this in above equation, we get


IF=elog(sin x)


But elog x=x, so he above equation becomes,


IF=sinx


So the general solution of a differential equation, is


y(IF)=∫(Q×I.F)dx+C


Substituting the corresponding values in the above equation, we get


y(sin x)=∫(4x cosec x (sin x))dx+C


But we know, so the above equation becomes,



y(sin x)=∫(4x.1)dx+C


y(sin x)=∫(4x)dx+C


On integrating, we get






But given y = 0 when , now substituting these values in above equation, we get



But , so above equation becomes,





Now substituting this value in equation (i), we get



Hence this is the particular solution of the differential equation


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