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Chapter 4: Problem 3

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

Short Answer

Expert verified
In summary, for the two given differential equations: 1. The first equation is exact, and the solution is found in implicit form as: $$ \Psi(x, y) = \frac{1}{2} x^4 - \frac{1}{2}x^2y^2 + \frac{1}{2}y^4 + C = C' $$ 2. The second equation is also exact, and the solution is found in implicit form as: $$ \Psi(x, y) = x^3 - x^2 - xy + y^2 + y^3 + C = C' $$

Step by step solution

01

(i) Verify if the given equation is exact and find integrating factors if needed

To verify if the given equation is exact, we first need to check if the following condition holds: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ where \(M = (2x^3 - xy^2)\) and \(N = (2y^3 - x^2y)\). Calculating the partial derivatives, we get: $$ \frac{\partial M}{\partial y} = -2xy $$ and $$ \frac{\partial N}{\partial x} = -2xy . $$ Since both partial derivatives are equal, the given equation is exact, and we don't need to find an integrating factor.
02

(i) Integrate to find the solution

As we have confirmed that the equation is exact, we can integrate \(M\) w.r.t \(x\) and \(N\) w.r.t \(y\) to find the potential function \(\Psi(x, y)\): $$ \Psi(x, y) = \int M dx = \int (2x^3 - xy^2) dx = \frac{1}{2} x^4 - \frac{1}{2}x^2y^2 + h(y) $$ Now, to find \(h(y)\), we differentiate \(\Psi(x, y)\) w.r.t \(y\) and compare it with \(N\): $$ \frac{\partial \Psi}{\partial y} = -x^2y + h'(y) $$ Comparing with \(N = (2y^3 - x^2y)\), we get: $$ h'(y) = 2y^3 \Rightarrow h(y) = \frac{1}{2}y^4 + C $$ Thus, the solution is: $$ \Psi(x, y) = \frac{1}{2} x^4 - \frac{1}{2}x^2y^2 + \frac{1}{2}y^4 + C = C' $$
03

(ii) Verify if the given equation is exact and find integrating factors if needed

To verify if the given equation is exact, we first need to check if the following condition holds: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ where \(M = (3x^2 - 2x - y)\) and \(N = (2y - x + 3y^2)\). Calculating the partial derivatives, we get: $$ \frac{\partial M}{\partial y} = -1 $$ and $$ \frac{\partial N}{\partial x} = -1 . $$ Since both partial derivatives are equal, the given equation is exact, and we don't need to find an integrating factor.
04

(ii) Integrate to find the solution

As we have confirmed that the equation is exact, we can integrate \(M\) w.r.t \(x\) and \(N\) w.r.t \(y\) to find the potential function \(\Psi(x, y)\): $$ \Psi(x, y) = \int M dx = \int (3x^2 - 2x - y) dx = x^3 - x^2 - xy + h(y) $$ Now, to find \(h(y)\), we differentiate \(\Psi(x, y)\) w.r.t \(y\) and compare it with \(N\): $$ \frac{\partial \Psi}{\partial y} = -x + h'(y) $$ Comparing with \(N = (2y - x + 3y^2)\), we get: $$ h'(y) = 2y + 3y^2 \Rightarrow h(y) = y^2 + y^3 + C $$ Thus, the solution is: $$ \Psi(x, y) = x^3 - x^2 - xy + y^2 + y^3 + C = C' $$

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