# Solve the equation (2xy − 3)dx + (x 2 + 4y)dy = 0 given that y(2) = 5.??

${x}^{2} y + 2 {y}^{2} - 3 x - 64 = 0$

#### Explanation:

Given DE

$\left(2 x y - 3\right) \mathrm{dx} + \left({x}^{2} + 4 y\right) \mathrm{dy} = 0$

Comparing above equation with the standard form of DE $M \mathrm{dx} + N \mathrm{dy} = 0$ we get

$M = 2 x y - 3 \setminus \implies \setminus \frac{\setminus \partial M}{\setminus \partial y} = 2 x$ &

$N = {x}^{2} + 4 y \setminus \implies \setminus \frac{\setminus \partial N}{\setminus \partial x} = 2 x$

Since, $\setminus \frac{\setminus \partial M}{\setminus \partial y} = \setminus \frac{\setminus \partial N}{\setminus \partial x}$ hence the given DE is an exact DE whose solution is given as

${\int}_{y = c o n s t} M \mathrm{dx} + \setminus {\int}_{\setminus \textrm{\mathfrak{e} e o f x}} N \mathrm{dy} = C$

$\setminus \int \left(2 x y - 3\right) \setminus \mathrm{dx} + \setminus \int \left({x}^{2} + 4 y\right) \setminus \mathrm{dy} = C$

$2 y \setminus \int x \setminus \mathrm{dx} - 3 \setminus \int \mathrm{dx} + 4 \setminus \int y \setminus \mathrm{dy} = C$

${x}^{2} y - 3 x + 2 {y}^{2} = C$

Now, applying initial condition, $y \left(2\right) = 5$ by setting $x = 2$ & $y = 5$ in above solution we get

${2}^{2} \left(5\right) - 3 \left(2\right) + 2 {\left(5\right)}^{2} = C$

$C = 64$

${x}^{2} y - 3 x + 2 {y}^{2} = 64$

${x}^{2} y + 2 {y}^{2} - 3 x - 64 = 0$