Question

Solve the following differential equation: (y – sin²x)dx + tanx dy = 0

Answer 1

Given differential equation is (y – sin²x)dx + tanx dy = 0

(y – sin²x)dx = –tanx dy

`(dy)/(dx) = (y - sin^2x)/(-tanx)`

`(dy)/(dx) = (sin^2x - y)/tanx`

`(dy)/(dx) = sin^2x/tanx - y/tanx`

`(dy)/(dx) = sinxcosx  –  ycotx`

`(dy)/(dx) + ycotx = sinxcosx`

Which is a linear differential equation of the form `(dy)/(dx) + Py = Q`

Where P = cot x

Q = sin x cos x

Here, If = `e^(intPdx) = e^(int cot xdx)`

= `e^(log|sinx|)` = sinx

∴ Solution is given by y.If = `int Q.If dx + C_1`

y.sinx = `int (sinx cosx sinx)dx + C_1`

y.sinx = `int sin^2x cosx dx + C_1`

y.sinx = I + C₁ ...(i)

Where I = `int sin^2x cosdx`

Let sinx = t

⇒ cosxdx = dt

∴  I = `int t^2dt = t^3/3 + C_2`

or I = `sin^3x/3 + C_2`

From equation (i), we have

y.sin = `sin^3x/3 + C_2 + C_1`

or y.sinx = `sin^3x/3 + C`  ...(Where C = C₁ + C₂)