Review_for_Exam__3

# Review_for_Exam__3 - LBS 119 EXAM#3 Review Topics 1...

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LBS 119 -- EXAM #3 – Review Topics 1. SEPARABLE EQUATIONS (1 st order): Arrange the equation so that dy dx is on the left side of the equation by itself. Separate the x’s and y’s on the other side by factoring out to get ( ) ( ) dy f x g y dx = . Bring the y’s to the left side and x’s to the right side to get 1 ( ) ( ) dy f x dx g y = . Integrate both sides of the equation. If possible, solve for y. 2. LINEAR EQUATIONS (1 st order): Put in standard form: ( ) ( ) dy P x y Q x dx + = . Find the integrating factor: ( ) P x dx e μ = Multiply the differential equation by the integrating factor: ( ) ( ) dy P x y Q x dx + = The left side becomes [ ] y so we get [ ] ( ) y Q x = Integrate both sides to get ( ) y Q x dx C = + Divide through by to finish solving for y. 3. BERNOULLI EQUATIONS (1 st order): Put in standard form: ( ) ( ) n dy P x y Q x y dx + = Let 1 (1 ) n n dv dy v y and n y dx dx - - = = - . Multiply the standard form equation by (1 ) n n y - - and substitute to get a standard form LINEAR equation with v’s instead of y’s. Finish solving for v with the LINEAR method. Substitute y back into the equation and solve for y if possible.

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4. EXACT EQUATIONS (Remember, these involve PARTIAL derivatives): Put into form ( , ) ( , ) 0 M x y dx N x y dy + = and CHECK for EXACTNESS: M N y x x = If exact, then there exists a function ( , ) f x y such that f f M and N x y x = = (this is by definition). Start with either one of these facts and work through to find ( , ) f x y using partial integrals and partial derivatives. Remember to be extra careful when keeping track of which variable is considered a variable and which variable is considered a constant in each step. The general solution is ( , ) f x y C = 5. HOMOGENEOUS LINEAR EQUATIONS (2 nd order or higher): Example: ay” + by’ + cy =0. Use a characteristic equation to solve these. An n th order equation needs n linearly independent parts to its general solution. Remember, to have linear independence, no two parts of the solution can be constant multiples of each other. If the characteristic equation has DISTINCT REAL ROOTS r 1 and r 2 , the general solution is 1 2 1 2 = + r x r x c y C e C e REPEATED REAL ROOTS r 1 = r 2 , the general solution is 1 1 1 2 = + r x r x c y C e C xe COMPLEX or IMAGINARY ROOTS 1 2 α β = + = - r i and r i , the general solution is 1 2 cos sin = + x x c y C e x C e x 6. NON-HOMOGENEOUS LINEAR EQUATIONS (2 nd order or higher): Example: y” + by’ + cy = g(x). (make sure there is no leading coefficient!)
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Review_for_Exam__3 - LBS 119 EXAM#3 Review Topics 1...

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