Solving First-Order Differential and Difference Equations
Learn how to solve first-order differential and difference equations step by step, including finding homogenous and particular solutions. Understand the process with practical examples and clear explanations.
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First-Order Differential Equation General Form: Step 1: Solve the Homogenous Part c c c c + = + = 0 0 0 y y y y 1 1 1 2
First-Order Difference Equation = A e h t t Y Step 2: Solve for the non-Homogenous Part (Particular Solution) ) First Case ( ) g t = (Where is a constant) 3
First-Order Difference Equation = ( ) y t y = Guess: 0 + = = = 0 c c c 1 0 0 c 0 c 0 In order for to be a solution we need 0 c = 0 If by t our guess. then this is not a solution and we have to multiply 0 = ( ) y t y = t New Guess: 4
First-Order Difference Equation B) Second Case ( ) is a generic function of time g t = h t ( ) '( ) Y y A t A t e e t = + t t ( ) ( ) A t e : y ? 1 c 1 c c c c c y y y + = + = + = ( ) ( ) ( ) 0 c c y g t y g t y g t 1 1 0 1 1 1 1 5
First-Order Difference Equation 1 c y + = + + = t t t ( ) '( ) ( ) ( ) ( ) ( ) y g t A t e A t e A t e G t 1 = = t t '( ) ( ) '( ) ( ) A t e G t A t e G t = t ( ) ( ) dA t e G t dt = + t ( ) ( ) A t e G t dt B 1 c = ( ) ( ) G t g t : 1 6
First-Order Difference Equation = = + GS t t t ( ) ( ) Y A t e e G t dt B e t = B e + GS t t t ( ) Y e e G t dt t hom ogenous particular 7
First-Order Difference Equation Step 3: Find the value of B = B e + GS t t t ( ) Y e e G t dt t hom ogenous particular In order to find B we need an initial condition. If we know that for, we substitute the initial condition into the general solution and we take B. = = * * * ( ) y t t t y 8
1 y + = 2 t 3 6 y(0)=1 y e 1 3 y y + = + = 2 2 t t 3 6 2 y e y e Step 1: Homogenous Solution y + = = = 2 h t h t 2 0 ( ) ( ) y Y A t e Y A t e t t 9
1 Step 2: Particular Solution = 2 h t ( ) '( ) Y y A t A t e e t = 2 2 t t 2 ( ) A t e 1 3 1 3 y + = + = 2 2 2 2 2 t t t t t 2 '( ) 2 ( ) 2 ( ) y e A t e A t e A t e e 1 3 1 3 = = 2 2 t t '( ) A t e e 10
1 1 3 1 3 1 3 = = = + 2 2 t t '( ) ( ) A t e e A t t d 1 3 = = + 2 2 GS t t ( ) t Y A t e t d e So: Step 3: Find value of d 1 3 2 0 = + = = (0) 1 0 1 1 y d e d 1 3 e = + 2 GS t 1 Y t t 11
2 y+ = + 2 2 y t Step 1: Homogenous Solution y + = = = 1 h t h t 0 ( ) ( ) y Y A t e Y A t e t t Step 2: Particular Solution = 1 h t ( ) Y y A t e t = 1 1 t t '( ) 1 ( ) A t e A t e 12
2 y + = + + = + 1 1 1 t t t 2 2 '( ) 1 ( ) ( ) 2 2 y t A t e A t e A t e t ( ) = + t '( ) 2 2 A t t e : = = ( ) ( ) f x g x dx ( ) ( ( )) f x d G x ( ) ( ) f x G x ( ) ' G x f ( ) x dx + = + = + + = t t t t (2 2) (2 2)( )' e (2 2) (2 2)' t e dt t dt t e e t dx = + = + = + = t t t t t t t t (2 2) 2 (2 2) 2 2 2 2 2 t e e dt t e e te e e te 13
2 = + t ( ) 2 A t te c = = + = + 1 GS t t t t ( ) 2 2 tY A t e te c e t c e Step 3: Find value of c = + = = 0 (0) 1 2 0 1 1 y c e c e = + GS t 2 t y t : 14
Second-Order Difference Equation General Form: y y + + = ( ) c c c y g t 2 1 0 Step 1: Solve the Homogenous Part c c c c c c y y y y + + = + + = 0 0 0 y y 2 1 1 2 2 2 2 15
Second-Order Difference Equation e = t ty Guess: : + + = 2 0 1 2 The solution depends on the roots = = 2 1 4 1 1) If >0 2 1,2 2 = + t t h ty A e A e 1 2 So, the solution is: 1 2 16
Second-Order Difference Equation = = = 2 1 * 4 1 2) If =0 2 1,2 2 e t second-order t e t . = + A t e h t t ty A e So, the solution is: 1 2 17
Second-Order Difference Equation = = 2 1 4 a i 3) If <0 2 1,2 1 2 = 1 1[4 2 = 2 1 ] 2 = + h t [ cos( ) sin( )] ty e A t A t So, the solution is: 1 2 18
1 y y + = 6 9 0 y Step 1: Homogenous Solution y y + = + = = 2 6 9 0 6 9 0 3 y = = + A t e 3 3 GS t h t t t y y A e 1 2 19
