Question

(y^2 e^xy^2 + 4x^3) dx + (2xye^xy^2 - 3y^2) dy = 0

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The solution of the given differential equation is $y = \frac{2e^{xy^{2}}}{x^{2}} + x^{4} - y^{3} + C$

Step-by-step explanation:

Given: The differential equation $(y^2 e^{xy^2}+ 4x^3) dx + (2xye^{xy^2} - 3y^2) dy = 0$

To Find: Solution of the given differential equation

Solution:

• Finding Solution of the given differential equation

• We have given $(y^2 e^{xy^2}+ 4x^3) dx + (2xye^{xy^2} - 3y^2) dy = 0$ such that it is in the form of $Mdx + Ndy = 0$.

• The differential equation is exact if it satisfies the following condition, $\Big( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \Big)$

For the given differential equation, we have $M = (y^2 e^{xy^2}+ 4x^3)$ and $N = (2xye^{xy^2} - 3y^2)$

Therefore, $\frac{\partial M}{\partial y} = 2ye^{xy^2}+ y^2 (2xye^{xy^2})$ and $\frac{\partial N}{\partial x} = 2ye^{xy^2}+ y^2 (2xye^{xy^2})$, which implies $\Big( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \Big)$, the differential equation is exact and can be solved in the following way,

$y = (\int{M} dx )_{\text{keeping y constant}} + (\int{N} dy )_{\text{not considering the expressions with x}}$

$\Rightarrow y = \int{(y^{2} e^{xy^2} + 4x^{3})} dx + \int{(-3y^{2} )}$

$\Rightarrow y = \frac{2e^{xy^{2}}}{x^{2}} + x^{4} - y^{3} + C$

Hence, the solution of the differential equation $(y^2 e^{xy^2}+ 4x^3) dx + (2xye^{xy^2} - 3y^2) dy = 0$ is $y = \frac{2e^{xy^{2}}}{x^{2}} + x^{4} - y^{3} + C$