## Exact first order ordinary differential equation

Solve the following ODE:
$$
x\frac{\mathrm d y}{\mathrm d x} + 5x + y = 0
$$

Solution:

The above equation can be written as $$ (5x+y)\ \mathrm d x + x\ \mathrm d y = 0 \tag{1} $$ Equation \( (1)\) is in the form of $$ A(x,y)\ \mathrm d x + B(x,y)\ \mathrm d y = 0 \tag{2} $$ where $$ \begin{align} \\& A(x,y) = 5x + y \\& \\& B(x,y) = x \end{align} $$

Let us say \( U(x,y) = c \ \) is the solution the equation \( (2), \ c \) is a constant.

So $$ \begin{align} \\& \mathrm d U = 0 \\& \Rightarrow \frac{\partial U}{\partial x}\ \mathrm d x + \frac{\partial U}{\partial y}\ \mathrm d y = 0 \tag{3} \end{align} $$ Comparing equation \( (2) \) and equation \( (3) \) we get $$ \begin{align} \\& A(x,y) = \frac{\partial U}{\partial x} \tag{4} \\& \\& B(x,y) = \frac{\partial U}{\partial y} \tag{5} \end{align} $$ Since \( \frac{\partial^2 U}{\partial x \partial y} = \frac{\partial^2 U}{\partial y \partial x}\) we therefore require $$ \boxed{ \frac{\partial A}{\partial y} = \frac{\partial B}{\partial x} } $$ Which is the required condition for equation \( (2) \) to be an exact equation.

Now $$ \begin{align} \\& A(x,y) = 5x + y \\& \Rightarrow \frac{\partial A}{\partial y} = 1 \\& \\& B(x,y) = x \\& \Rightarrow \frac{\partial B}{\partial x} = 1 \end{align} $$

Which satisfies the required condition so equation \( (1) \) is an exact equation. Now from equation \( (4) \): $$ \begin{align} \\& \frac{\partial U}{\partial x} = A(x,y) = 5x+y \\& \Rightarrow U(x,y) = \frac{5x^2}{2} + yx + \phi(y) \tag{6} \end{align} $$ Here \( \phi(y) \) is a function of \( y \) only. Remember \( U(x,y) \) is a function of \(x \) and \(y\). So we cannot put an ordinary constant there. That is why a function \( \phi(y)\). If you take the partial derivative of the equation (6) with respect to \( x\) you will get the previous expression back.

Now take the partial derivative of equation (6) with respect to \( y\) and equate with the equation (5): $$ \begin{align} \\& U(x,y) = \frac{5x^2}{2} + yx + \phi(y) \\& \Rightarrow \frac{\partial U}{\partial y} = B(x,y) = 0 + x + \frac{\mathrm d \phi}{\mathrm d y} \tag{7} \\& \Rightarrow x+\frac{\mathrm d \phi}{\mathrm d y} = x \\& \Rightarrow \frac{\mathrm d \phi}{\mathrm d y} = 0 \\& \Rightarrow \phi(y) = k \end{align} $$ \( k\) is a constant. So the complete solution of the equation (1) is $$ \begin{align} \\& U(x,y) = c \\& \Rightarrow \frac{5x^2}{2} + yx + \phi(y) = c \\& \Rightarrow \frac{5x^2}{2} + yx + k = c \\& \Rightarrow \frac{5x^2}{2} + yx = c - k =c_1 \\& \\& \boxed{ \frac{5x^2}{2} + yx = c_1 } \end{align} $$ \( c_1\) is a constant.

Solution:

The above equation can be written as $$ (5x+y)\ \mathrm d x + x\ \mathrm d y = 0 \tag{1} $$ Equation \( (1)\) is in the form of $$ A(x,y)\ \mathrm d x + B(x,y)\ \mathrm d y = 0 \tag{2} $$ where $$ \begin{align} \\& A(x,y) = 5x + y \\& \\& B(x,y) = x \end{align} $$

Let us say \( U(x,y) = c \ \) is the solution the equation \( (2), \ c \) is a constant.

So $$ \begin{align} \\& \mathrm d U = 0 \\& \Rightarrow \frac{\partial U}{\partial x}\ \mathrm d x + \frac{\partial U}{\partial y}\ \mathrm d y = 0 \tag{3} \end{align} $$ Comparing equation \( (2) \) and equation \( (3) \) we get $$ \begin{align} \\& A(x,y) = \frac{\partial U}{\partial x} \tag{4} \\& \\& B(x,y) = \frac{\partial U}{\partial y} \tag{5} \end{align} $$ Since \( \frac{\partial^2 U}{\partial x \partial y} = \frac{\partial^2 U}{\partial y \partial x}\) we therefore require $$ \boxed{ \frac{\partial A}{\partial y} = \frac{\partial B}{\partial x} } $$ Which is the required condition for equation \( (2) \) to be an exact equation.

Now $$ \begin{align} \\& A(x,y) = 5x + y \\& \Rightarrow \frac{\partial A}{\partial y} = 1 \\& \\& B(x,y) = x \\& \Rightarrow \frac{\partial B}{\partial x} = 1 \end{align} $$

Which satisfies the required condition so equation \( (1) \) is an exact equation. Now from equation \( (4) \): $$ \begin{align} \\& \frac{\partial U}{\partial x} = A(x,y) = 5x+y \\& \Rightarrow U(x,y) = \frac{5x^2}{2} + yx + \phi(y) \tag{6} \end{align} $$ Here \( \phi(y) \) is a function of \( y \) only. Remember \( U(x,y) \) is a function of \(x \) and \(y\). So we cannot put an ordinary constant there. That is why a function \( \phi(y)\). If you take the partial derivative of the equation (6) with respect to \( x\) you will get the previous expression back.

Now take the partial derivative of equation (6) with respect to \( y\) and equate with the equation (5): $$ \begin{align} \\& U(x,y) = \frac{5x^2}{2} + yx + \phi(y) \\& \Rightarrow \frac{\partial U}{\partial y} = B(x,y) = 0 + x + \frac{\mathrm d \phi}{\mathrm d y} \tag{7} \\& \Rightarrow x+\frac{\mathrm d \phi}{\mathrm d y} = x \\& \Rightarrow \frac{\mathrm d \phi}{\mathrm d y} = 0 \\& \Rightarrow \phi(y) = k \end{align} $$ \( k\) is a constant. So the complete solution of the equation (1) is $$ \begin{align} \\& U(x,y) = c \\& \Rightarrow \frac{5x^2}{2} + yx + \phi(y) = c \\& \Rightarrow \frac{5x^2}{2} + yx + k = c \\& \Rightarrow \frac{5x^2}{2} + yx = c - k =c_1 \\& \\& \boxed{ \frac{5x^2}{2} + yx = c_1 } \end{align} $$ \( c_1\) is a constant.