Question

Solve the following differential equations : (i) \(\left(2 x \cos y+y^{2} \cos x\right) d x\) \(+\left(2 y \sin x-x^{2} \sin y\right) d y=0\) (ii) \(\frac{x^{3} d x+y x^{2} d y}{\sqrt{x^{2}+y^{2}}}=y d x-x d y\)

Short answer

Question: Solve the following differential equations: (i) \(\left(2 x \cos y+y^{2} \cos x\right) d x+\left(2 y \sin x-x^{2} \sin y\right) d y=0\) (ii) \(\frac{x^{3} d x+y x^{2} d y}{\sqrt{x^{2}+y^{2}}}=y d x-x d y\) Answer: (i) The general solution for the first differential equation is: \(F(x, y) = 2x^2 \cos y + y^2 x \cos x + C = 0\) (ii) The general solution for the second differential equation is: \(\dfrac{y}{x}+\dfrac{2}{3}\left(\dfrac{y}{x}\right)^3+\dfrac{1}{5}\left(\dfrac{y}{x}\right)^5 = \dfrac{1}{2}x^2 + C\)

Step-by-step solution

Classify differential equation (i)

The given first differential equation is: \(\left(2 x \cos y+y^{2} \cos x\right) d x+\left(2 y \sin x-x^{2} \sin y\right) d y=0\) This is an exact differential equation in the form \(M dx + N dy = 0\), where \(M = 2x \cos y + y^2 \cos x\) and \(N = 2y \sin x - x^2 \sin y\). We can check if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\): \(\frac{\partial M}{\partial y} = -2x \sin y + 2y \cos x\) \(\frac{\partial N}{\partial x} = 2y \sin x - 2x \sin y\) Comparing the partial derivatives, we confirm that the equation is indeed exact.

Solve the exact differential equation (i)

Since the equation is exact, we can find a potential function \(F(x, y)\) such that: \(\frac{\partial F}{\partial x} = M = 2x \cos y + y^2 \cos x\) Integrating with respect to x, we get: \(F(x, y) = \int M dx = \int (2x \cos y + y^2 \cos x) dx = 2x^2 \cos y + y^2 x \cos x + g(y)\) Now, we differentiate the function with respect to y: \(\frac{\partial F}{\partial y} = -2x^2 \sin y + 2yx \cos x + x^2 y \sin x + g'(y)\) We also know that: \(\frac{\partial F}{\partial y} = N = 2y \sin x - x^2 \sin y\) Comparing, we get: \(g'(y) = 0 \Rightarrow g(y) = C\) The general solution is: \(F(x, y) = 2x^2 \cos y + y^2 x \cos x + C = 0\)

Classify differential equation (ii)

The given second differential equation is: \(\frac{x^{3} d x+y x^{2} d y}{\sqrt{x^{2}+y^{2}}}=y d x-x d y\) This is a homogeneous equation because it is in the form: \(\frac{dy}{dx} = \frac{F(tx, ty)}{G(tx, ty)}\) Dividing both sides by \(dx\) and rearranging the terms, we obtain: \(H\dfrac{dy}{dx} - x = K\dfrac{y}{x}\) Where \(K = \dfrac{x^3}{(x^2+y^2)^\frac{1}{2}}\) and \(H = \dfrac{y}{(x^2+y^2)^\frac{1}{2}}\). Now, substitute \(v = \frac{y}{x}\) and \(y = v(x)x\): \((1 + v\dfrac{dv}{dx})x = Kv\)

Solve the homogeneous differential equation (ii)

The homogeneous equation for (ii) becomes: \(\left(1+\dfrac{y}{x}\dfrac{dv}{dx}\right) x = \frac{x^3}{\sqrt{x^2+y^2}} v\) Substituting \(v=\dfrac{y}{x}\), we get: \(1+\dfrac{y}{x} \,\dfrac{dv}{dx} = \dfrac{v}{\sqrt{1+v^2}}\) Now we solve the separable equation by dividing the LHS by \(1+v^2\) and the RHS by \(v\): \(\dfrac{1+\dfrac{y}{x}\dfrac{dv}{dx}}{1+\left(\dfrac{y}{x}\right)^2} = \dfrac{1}{\sqrt{1+\left(\dfrac{y}{x}\right)^2}}\) Let \(z = \dfrac{y}{x} \Rightarrow \dfrac{dz}{dx} = \dfrac{dv}{dx}\): \(\dfrac{1+z \,\dfrac{dz}{dx}}{1+z^2} = \dfrac{1}{\sqrt{1+z^2}}\) Separating variables: \((1+z^2)^2 \,dz = x\,dx\) Now, integrating both sides: \(\int (1+z^2)^2 \,dz = \int x\,dx\) Integrating the LHS we get: \(\int(1+2z^2+z^4)dz= z+\frac{2}{3}z^3+\frac{1}{5}z^5=C_1\) Integrating the RHS we get: \(\int x \, dx = \frac{1}{2}x^2 + C_2\) The general solution can be expressed as: \(z+\dfrac{2}{3}z^3+\dfrac{1}{5}z^5 = \dfrac{1}{2}x^2 + C\) Where \(C = C_1 - C_2\). Substituting back \(z = \dfrac{y}{x}\), we get the final general solution of the second differential equation: \(\dfrac{y}{x}+\dfrac{2}{3}\left(\dfrac{y}{x}\right)^3+\dfrac{1}{5}\left(\dfrac{y}{x}\right)^5 = \dfrac{1}{2}x^2 + C\)