$ysin(2x)-y + 2y^{2}e^{xy^{2}}dx=x-sin^{2}(x)-4xye^{xy^{2}}dy$
$ysin(2x)-y + 2y^{2}e^{xy^{2}}dx=x-sin^{2}(x)-4xye^{xy^{2}}dy$
Answer
Explanation:
Step1: Rearrange the differential equation
Rewrite the given equation $[y\sin(2x)-y + 2y^{2}e^{xy^{2}}]dx=[x-\sin^{2}(x)-4xye^{xy^{2}}]dy$ as ((y\sin(2x)-y + 2y^{2}e^{xy^{2}})dx+( -x+\sin^{2}(x)+4xye^{xy^{2}})dy = 0). Let (M=y\sin(2x)-y + 2y^{2}e^{xy^{2}}) and (N=-x+\sin^{2}(x)+4xye^{xy^{2}}).
Step2: Check exactness
Calculate (\frac{\partial M}{\partial y}=\sin(2x)-1 + 4ye^{xy^{2}}+4xy^{3}e^{xy^{2}}) and (\frac{\partial N}{\partial x}=-1 + 2\sin(x)\cos(x)+4ye^{xy^{2}}+4xy^{3}e^{xy^{2}}). Since (\sin(2x)=2\sin(x)\cos(x)), (\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}), so the differential - equation is exact.
Step3: Find the solution
We know that if the equation (Mdx + Ndy = 0) is exact, then (\int_{x_0}^{x}M(x,y_0)dx+\int_{y_0}^{y}N(x,y)dy = C) (or we can find a function (f(x,y)) such that (\frac{\partial f}{\partial x}=M) and (\frac{\partial f}{\partial y}=N)). Integrate (M) with respect to (x): (\int(y\sin(2x)-y + 2y^{2}e^{xy^{2}})dx=-\frac{y}{2}\cos(2x)-yx + 2e^{xy^{2}}+h(y)). Now differentiate this with respect to (y): (\frac{\partial f}{\partial y}=-\frac{1}{2}\cos(2x)-x + 4xye^{xy^{2}}+h^\prime(y)). Since (\frac{\partial f}{\partial y}=N=-x+\sin^{2}(x)+4xye^{xy^{2}}), and (\sin^{2}(x)=\frac{1 - \cos(2x)}{2}), we have (h^\prime(y) = 0). So the solution of the differential equation is (-\frac{y}{2}\cos(2x)-yx + 2e^{xy^{2}}=C).
Answer:
(-\frac{y}{2}\cos(2x)-yx + 2e^{xy^{2}}=C)