Find the particular solution of the differential equation given that y = 0 when

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=e^{log(sin x)}

But e^{log 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