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 Rate this question :

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