widdonod1t

Answered

2021-12-29

Solve the following IVP for the exact equation and find the interval of validity for the solution.

$2xy-9{x}^{2}+(2y+{x}^{2}+1)\text{}\frac{dy}{dx}=0\text{}y\left(0\right)=-2$

Answer & Explanation

Barbara Meeker

Expert

2021-12-30Added 38 answers

For any differential equation that is given to be exact, we must have a function f(x,y) such that:

$d\left[f(x,y)\right]=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=M\text{}dx+N\text{}dy$

$\therefore M(x,y)=\frac{\partial f}{\partial x}\text{}\text{and}\text{}N(x,y)=\frac{\partial f}{\partial y}$

$i.e.f(x,y)=\int M(x,y)dx+{\varphi}_{1}\left(y\right)\text{}\text{and}\text{}f(x,y)=\int N(x,y)dy+{\varphi}_{2}\left(x\right)$

The given differential equation can be written as:

$2xy-9{x}^{2}+(2y+{x}^{2}+1)\frac{dy}{dx}=0$

$(2xy-9{x}^{2})dx+(2y+{x}^{2}+1)dy=0$ (i)

Comparing to the standard form of$M(x,y)dx+N(x.y)dy=0$ we get:

$M(x,y)=2xy-9{x}^{2}$

$\therefore f(x,y)=\int (2xy-9{x}^{2})dx+{\varphi}_{1}\left(y\right)$

$=2y\left(\frac{{x}^{2}}{2}\right)-9\left(\frac{{x}^{3}}{3}\right)+{\varphi}_{1}\left(y\right)$

$={x}^{2}y-3{x}^{3}+{\varphi}_{1}\left(y\right)$ ...(2)

$N(x,y)=2y+{x}^{2}+1$

$\therefore f(x,y)=\int (2y+{x}^{2}+1)dy+{\varphi}_{2}\left(x\right)$

$=2\left(\frac{{y}^{2}}{2}\right)+{x}^{2}\left(y\right)+\left(y\right)+{\varphi}_{2}\left(x\right)$

The given differential equation can be written as:

Comparing to the standard form of

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reinosodairyshm

Expert

2021-12-31Added 36 answers

karton

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2022-01-09Added 439 answers

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