Education • All: 1.2: Initial Value Theorem

INITIAL VALUE THEOREM

$\large \\IF \;f(x,y)\;and\;\frac{\partial y}{\partial x}\;are\;continuous\;on\;(a,b) \\\\THEN\;there\;exist\;I_{0}\;(x_{0}-h,x_{0}+h),h>0 \\\\WHERE\;there's\;a\;unique\;fuction\;y(x)\;solution$

Example (1): When initial value is given

Is it xy'=2y unique when y(0) = 0?
$\large \fn_cm \\xy'=2y \\\\y'=\frac{2y}{x} \\\\\frac{dy}{dx}=\frac{2y}{x} \\\\\int\frac{dy}{2y}=\int\frac{dx}{x} \\\\\frac{1}{2}\ln y=\ln x \\\\\ln y^{\frac{1}{2}}=\ln x+C \\y^{\frac{1}{2}}=x \\y=x^2 \\y'=2x$
No! Equation has more than 1 solution (y=0 and y'=2x); Therefore it fails the test of uniqueness.

Example (2): No initial value is given

1. Obtain the domain of the given function.
2. Obtain the domain of the partial derivate of the function.
3. Combine both domains to obtain a final domain and see where there function is not continuous.

$\large \fn_cm \\\frac{dy}{dx}-y=x\;\;\;\;(Domain=(- \infty,\infty )) \\\\\frac{\partial f}{\partial y}=x+1 \\\\\frac{\partial f}{\partial y}=1\;\;\;\;(Domain(- \infty,\infty))$
Function is continues everywhere.

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1.2: Initial Value Theorem

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