I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form ofy′+P(x)y=Q(x)Here is the equation:(2xy+x2+x4)dx−(1+x2)dy=0It is not exact since partial derivatives are not equal.Any help would be appreciated.

To put it into the form you requested: −(1+x2)dy+(2xy+x2+x4)dx=0⇒ ⇒dydx−2xy+x2+x41+x2=0 ⇒dydx+(−2x1+x2)y=x2+x41+x2 ⇒dydx+(−2x1+x2)y=x2

We can write your equation as y′−2xyx2+1=x2

To put it into the form you requested: −(1+x2)dy+(2xy+x2+x4)dx=0⇒ ⇒dydx−2xy+x2+x41+x2=0 ⇒dydx+(−2x1+x2)y=x2+x41+x2 ⇒dydx+(−2x1+x2)y=x2

We can write your equation as y′−2xyx2+1=x2

Key to "I have no idea how to solve it...."

Tashan Flores

2022-02-24

I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of

y^{'} + P (x) y = Q (x)

Here is the equation:

(2 x y + x^{2} + x^{4}) d x - (1 + x^{2}) d y = 0

It is not exact since partial derivatives are not equal.
Any help would be appreciated.

tardanetkd2

Beginner2022-02-25Added 9 answers

To put it into the form you requested:

- (1 + x^{2}) d y + (2 x y + x^{2} + x^{4}) d x = 0 \Rightarrow

\Rightarrow \frac{d y}{d x} - \frac{2 x y + x^{2} + x^{4}}{1 + x^{2}} = 0

\Rightarrow \frac{d y}{d x} + (- \frac{2 x}{1 + x^{2}}) y = \frac{x^{2} + x^{4}}{1 + x^{2}}

\Rightarrow \frac{d y}{d x} + (- \frac{2 x}{1 + x^{2}}) y = x^{2}

Tommie Bryan

Beginner2022-02-26Added 4 answers

We can write your equation as

y^{'} - \frac{2 x y}{x^{2} + 1} = x^{2}

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