Answer 1

(x + y) dy = a² dx

∴ `dy/dx = a^2/(x+y)` ...(i)

Put x + y = t  ...(ii)

∴ y = t - x

Differentiating w.r.t. x, we get

∴ `dy/dx = dt /dx -1` ....(iii)

Substituting (ii) and (iii) in (i), we get

`dt/dx -1 = a^2/t`

∴ `dt/dx = a^2/t + 1`

∴ `dt/dx = (a^2+t)/t`

∴ `t/(a^2+t)  dt = dx`

Integrating on both sides, we get

`int ((a^2+t) - a^2)/(a^2+ t)  dt = int dx`

∴ `int 1 dt- a^2int 1/(a^2+t) dt = int dx`

∴ t - a² log |a² + t| = x + c₁

∴ x + y - a² log |a² + x + y| = x + c₁

∴ y - a² log |a² + x + y| = c₁

∴ y - c₁ = a² log |a² + x + y|

∴ `y/a^2 - c_1/a^2 = log |a^2 + x + y|`

∴ `a^2 + x + y = e^(a^(y/2). e^(a^((-c1)/2)`

∴ `a^2 + x + y = ce^(a^(y/2) ` … `[ c =e^(a^((-c1)/2)]]`