Question

$ysin(2x)-y + 2y^{2}e^{xy^{2}}dx=x-sin^{2}(x)-4xye^{xy^{2}}dy$

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:

\(-\frac{y}{2}\cos(2x)-yx + 2e^{xy^{2}}=C\)