Chapter 8: Q16P (page 407)

Solve the differential equation $y y^{2} + 2 x y' - y = 0$ by changing from variables role="math" localid="1655272385100" $y, x$ to $r, x$ where $y^{2} = r^{2}$ ; then $y y' = n' - x$ .

Short Answer

Expert verified

Answer

The solution of the differential equation is $y = \pm \sqrt{- \frac{r^{2}}{2} (r^{2} - {(r r')}^{2})}$ .

Step by step solution

Given information

A differential equation $y y'^{2} + 2 x y' - y = 0$

, where $y^{2} = r^{2} - x^{2}$ and $y y' = r r' - x$

Homogenous function

Homogenous equation: A homogeneous function of x and y of degree n means the function can be written as $x^{n} f (\frac{y}{x})$

$P (x, y) d x + Q (x, y) d y = 0$

, where P and Q are homogeneous function of the same degree.

By the change of variable of homogeneous equation,

$y' = \frac{d y}{d x} = \frac{P (x, y)}{Q (x, y)} = f (\frac{y}{x})$

Substitute $y = x v$ in the equation.

Solve the differential equation by use of homogeneous function

Put the value of y' in the equation as,

$y {(\frac{r r' - x}{y})}^{2} + 2 x (\frac{r r' - x}{y}) - y = 0 \frac{1}{2} {(r r' - x)}^{2} + 2 x (\frac{r r' - x}{y}) - y = 0 \frac{1}{y} \{{(r r' - x)}^{2} + 2 x (r r' - x) - y^{2}\} = 0$

Put $y^{2} = r^{2} - x^{2}$

${(r r' - x)}^{2} + 2 x (r r' - x) - (r r' - x) - (r^{2} - x^{2}) = 0 - 2 x^{2} - r^{2} + {(r r')}^{2} = 0 . . . . . . (1)$

Replace by $y = x r$ ,

$\frac{y}{r} = x$

Put in equation (1) as,

$- 2 {(\frac{y}{r})}^{2} - r^{2} + {(r r')}^{2} = 0 - \frac{2 y^{2}}{r^{2}} - r^{2} + {(r r')}^{2} = 0 y = \pm \sqrt{- \frac{r^{2}}{2} (r^{2} - {(r r')}^{2})}$

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Solve the differential equation $y y^{2} + 2 x y' - y = 0$ by changing from variables role="math" localid="1655272385100" $y, x$ to $r, x$ where $y^{2} = r^{2}$ ; then $y y' = n' - x$ .

Short Answer

Step by step solution

Given information

Homogenous function

Solve the differential equation by use of homogeneous function

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Solve the differential equation yy2+2xy'-y=0by changing from variables role="math" localid="1655272385100" y,xto r,x where y2=r2; then yy'=n'-x.

Short Answer

Step by step solution

Given information

Homogenous function

Solve the differential equation by use of homogeneous function

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Oscillations

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Solve the differential equation $y y^{2} + 2 x y' - y = 0$ by changing from variables role="math" localid="1655272385100" $y, x$ to $r, x$ where $y^{2} = r^{2}$ ; then $y y' = n' - x$ .