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=EGM6321 - Principles of Engineering Analysis 1, Fall 2009= MEETING 5 - Thursday, 03Sept09

5-1
Note: p4-1 Eq(2) and Eq(4) $$ y'(x)=p(x) \ $$ Integrate from a to x: $$ \int_{s=a}^{s=x} y'(s)\, ds = \int_{s=a}^{s=x} p(s)\, ds $$ $$ \left [ y(s) \right ]_{s=a}^{s=x} = y(x)-y(a) $$, where $$ y(a) = constant \ $$, $$ y(x)=\int_{s=a}^{s=x} p(s)\, ds +y(a)= \int_{}^{x} p(s)\, ds $$ another way: $$ y(x)=\int_{}^{} p(x)\, dx +k $$ Where $$ \int_{}^{} p(x)\, dx = F(x) $$ and $$ k=constant \  $$ $$ \Rightarrow \ y(a)=F(a)+k \ $$ $$ \Rightarrow \ k=y(a)-F(a) \ $$ $$ \Rightarrow \ y(x)=F(x)-F(a)+y(a) \ $$

5-2
But $$ F(x)-F(a)=\int_{s=a}^{s=x} p(s)\, ds \ $$ EQ1) $$\Rightarrow \ y(x)=\int_{s=a}^{s=x} p(s)\, ds+y(a)=\int_{}^{x} p(s)\, ds=\int_{}^{} p(x)\, dx+k \ $$ p4-2 Eq(3): Why this form of nonlinear 1st order ODE? Most general form: EQ2) $$ F(x,y,y')=0 \ $$ Application: EQ3) $$ x^2y^5+6(y')^2=0 \ $$ Where $$ x^2y^5+6(y')^2 \ $$ is defined as $$ F(x,y,y') \ $$ HW: Show that $$ F(x,y,y')=0 \ $$ in Eq(3) is a nonlinear 1st order ODE. Hint: Define the differential operator $$ D(.) \ $$ associated with Eq(3).