(i) Given differential equation: [ x,dx + y,dy + 4y^3(x^2 + y^2),dy = 0 ] Combine (y,dy) terms: [ x,dx + y,dy(1 + 4y^2(x^2 + y^2)) = 0 ] Let us write: [ x,dx + y(1 + 4y^2(x^2 + y^2)),dy = 0 ] This can be rearranged as: [ \frac{dx}{dy} = -\frac{y(1 + 4y^2(x^2 + y^2))}{x} ] Notice the term (x^2 + y^2), suggest substitution (u = x^2 + y^2), then (du = 2x,dx + 2y,dy \Rightarrow x,dx + y,dy = \frac{1}{2} du). Then equation becomes: [ \frac{1}{2} du + 4y^3(x^2 + y^2),dy = \frac{1}{2} du + 4y^3 u,dy = 0 ] Multiply both sides by 2: [ du + 8 y^3 u,dy = 0 ] This is a linear differential equation in (u) with respect to (y): [ \frac{du}{dy} + 8y^3 u = 0 ] This is separable: [ \frac{du}{u} = -8y^3 dy ] Integrate both sides: [ \ln|u| = -2y^4 + C ] Exponentiate: [ u = C e^{-2y^4} \quad \text{where } C \text{ is constant} ] Recall (u = x^2 + y^2), so: [ x^2 + y^2 = C e^{-2y^4} ] This is the solution of the first equation. --- (ii) Given differential equation: [ y'' + 4y = 2\cos x \cos 3x ] Use the identity: (\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]) [ 2\cos x \cos 3x = 2 \cdot \frac{1}{2}[\cos(3x-x) + \cos(3x+x)] = \cos 2x + \cos 4x ] So equation becomes: [ y'' + 4y = \cos