Question

The solution of dy/dx + y = y²cosx is​

Answer 1

Answer:

dy/dx +ycosx = yⁿsin2x

Multiplying the whole equation by y-ⁿ

y-ⁿ(dy/dx) + y¹-ⁿcosx = sin2x ……(1)

The above is the Bernoulli's differential equation which is linear in y.

Put u=y¹-ⁿ

du/dx = (1-n)y-ⁿ(dy/dx)

So, (1/(1-n))(du/dx) = y-ⁿ(dy/dx)

Substitute in the eqn.(1)

(1/(1-n))(du/dx) + ucosx = sin2x

du/dx +(1-n)cosx.u=(1-n)sin2x

This is linear in u, so the solution is

IF =e^[ ∫(1-n)cosxdx]=e^[(1-n)sinx]

So, solution is

u(IF)= ∫Q(IF)dx + C

u[e^[(1-n)sinx]]= ∫[(1-n)sin2x.e^[(1-n)sinx]]dx + c

ue^[(1-n)sinx]=(1-n) ∫[e^[(1-n)sinx].sin2xdx + c

z= ∫[e^[(1-n)sinx]sin2xdx

sinx = t

Cosx dx =dt

I= ∫[e^[(1-n)t]2tdt

I=2[t ∫e^[(1-n)t]dt - ∫D(t) ∫e^[(1-n)t]dtdt]

I=2[t(e^[(1-n)t])/(1-n)]-2 ∫1.[[e^[(1-n)t]]/(1-n)]dt

I=[2te^[(1-n)t]/(1-n)] -(2/(1-n)) ∫e^[(1-n)t]dt

I=[2te^[(1-n)t]]/(1-n) - [2/(1-n)²]e^[(1-n)t] + c

I=[2sinxe^[(1-n)sinx]]/(1-n) - [2/(1-n)²]e^[(1-n)sinx] + c

So the solution is

ue^[(1-n)sinx]=(1-n) ∫[e^[(1-n)sinx].sin2xdx + c

ue^[(1-n)sinx]=(1-n)[2sinxe^[(1-n)sinx]]/(1-n) - [2/(1-n)²]e^[(1-n)sinx] + c

ue^[(1-n)sinx]=2sinx.e^[(1-n)sinx] - [2/(1-n)]e^[(1-n)sinx]+c

Multiply by e^[(n-1)sinx]

u=2sinx-(2/(1-n)) + ce^[(n-1)sinx]

But u=y¹-ⁿ

y¹-ⁿ= 2sinx +[2/(n-1)]+ ce^[(n-1)sinx]

Multiply by (n-1)

(n-1)y¹-ⁿ=2(n-1)sinx + 2 + c(n-1)e^[(n-1)sinx]