6. $left2y - \frac{1}{x}+cos(3x)\right\frac{dy}{dx}+\frac{y}{x^{2}}-4x^{3}+3ysin(3x)=0$
6. $left2y - \frac{1}{x}+cos(3x)\right\frac{dy}{dx}+\frac{y}{x^{2}}-4x^{3}+3ysin(3x)=0$
Answer
Explanation:
Step1: Rewrite the equation
We rewrite the given differential equation (\left(2y-\frac{1}{x}+\cos(3x)\right)\frac{dy}{dx}+\frac{y}{x^{2}}-4x^{3}+3y\sin(3x) = 0) as (\left(2y-\frac{1}{x}+\cos(3x)\right)dy+\left(\frac{y}{x^{2}}-4x^{3}+3y\sin(3x)\right)dx = 0). Let (M(x,y)=\frac{y}{x^{2}}-4x^{3}+3y\sin(3x)) and (N(x,y)=2y - \frac{1}{x}+\cos(3x)).
Step2: Check exactness
Calculate (\frac{\partial M}{\partial y}=\frac{1}{x^{2}}+3\sin(3x)) and (\frac{\partial N}{\partial x}=\frac{1}{x^{2}}-3\sin(3x)). Since (\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}), the equation is not exact. We try to find an integrating - factor.
Step3: Find integrating factor
We calculate (\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}=\frac{\left(\frac{1}{x^{2}} + 3\sin(3x)\right)-\left(\frac{1}{x^{2}}-3\sin(3x)\right)}{2y-\frac{1}{x}+\cos(3x)}=\frac{6\sin(3x)}{2y-\frac{1}{x}+\cos(3x)}), which is not a function of (x) alone. Calculate (\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=\frac{\left(\frac{1}{x^{2}}-3\sin(3x)\right)-\left(\frac{1}{x^{2}}+3\sin(3x)\right)}{\frac{y}{x^{2}}-4x^{3}+3y\sin(3x)}=\frac{- 6\sin(3x)}{\frac{y}{x^{2}}-4x^{3}+3y\sin(3x)}), which is not a function of (y) alone. Let's rewrite the equation as (\left(2y-\frac{1}{x}+\cos(3x)\right)dy=-\left(\frac{y}{x^{2}}-4x^{3}+3y\sin(3x)\right)dx). We can rewrite it in the form (\frac{dy}{dx}=-\frac{\frac{y}{x^{2}}-4x^{3}+3y\sin(3x)}{2y-\frac{1}{x}+\cos(3x)}). Let's try to separate variables. First, group the terms with (y) together: (\left(2y-\frac{1}{x}+\cos(3x)\right)dy+\left(-\frac{y}{x^{2}} + 4x^{3}-3y\sin(3x)\right)dx = 0) (2y\ dy-\frac{1}{x}dy+\cos(3x)dy-\frac{y}{x^{2}}dx + 4x^{3}dx-3y\sin(3x)dx = 0) (2y\ dy+\left(\cos(3x)dy-3y\sin(3x)dx\right)-\left(\frac{1}{x}dy+\frac{y}{x^{2}}dx\right)+4x^{3}dx = 0) We know that (d(y^{2}) = 2y\ dy), (d(y\cos(3x))=\cos(3x)dy-3y\sin(3x)dx) and (d\left(\frac{y}{x}\right)=\frac{1}{x}dy-\frac{y}{x^{2}}dx). The equation becomes (d(y^{2})+d(y\cos(3x))-d\left(\frac{y}{x}\right)+4x^{3}dx = 0).
Step4: Integrate
Integrating both sides: (\int d(y^{2})+\int d(y\cos(3x))-\int d\left(\frac{y}{x}\right)+\int4x^{3}dx = C) (y^{2}+y\cos(3x)-\frac{y}{x}+x^{4}=C)
Answer:
(y^{2}+y\cos(3x)-\frac{y}{x}+x^{4}=C)