Question

solve the d.e. (x - y^6 + y^2 sin(x)) dx=(6xy^5 + 2y cos(x)) dy

Explanation:

Step1: Rearrange the differential equation

We rewrite the given differential equation \((x - y^{6}+y^{2}\sin(x))dx=(6xy^{5}+2y\cos(x))dy\) as \((x - y^{6}+y^{2}\sin(x))dx-(6xy^{5}+2y\cos(x))dy = 0\). Let \(M(x,y)=x - y^{6}+y^{2}\sin(x)\) and \(N(x,y)=-(6xy^{5}+2y\cos(x))=- 6xy^{5}-2y\cos(x)\).

Step2: Check exactness

Calculate \(\frac{\partial M}{\partial y}=-6y^{5}+2y\sin(x)\) and \(\frac{\partial N}{\partial x}=-6y^{5}+2y\sin(x)\). Since \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\), the differential - equation is exact.

Step3: Find the potential function \(f(x,y)\)

Integrate \(M(x,y)\) with respect to \(x\): \(\int(x - y^{6}+y^{2}\sin(x))dx=\frac{x^{2}}{2}-y^{6}x - y^{2}\cos(x)+h(y)\).
Differentiate \(f(x,y)\) with respect to \(y\): \(\frac{\partial f}{\partial y}=-6xy^{5}-2y\cos(x)+h^{\prime}(y)\). Since \(\frac{\partial f}{\partial y}=N(x,y)=-6xy^{5}-2y\cos(x)\), then \(h^{\prime}(y) = 0\), so \(h(y)=C\) (a constant).

Step4: Write the solution

The solution of the exact differential equation is \(\frac{x^{2}}{2}-y^{6}x - y^{2}\cos(x)=C\).

Answer:

\(\frac{x^{2}}{2}-y^{6}x - y^{2}\cos(x)=C\)