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Unformatted text preview: MATH 150 Linear Equations and it’s Application on Compound interest Cheung Man Wai Tutorial 3 1 Linear Equation A first-order linear differential equation appears in the form dy dx + p ( x ) y = g ( x ) , with y ( x ) = y . Applying the so-called integrating factor, the solution to (2) is given by y = 1 μ ( x ) y + Z x x μ ( x ) g ( x ) dx , where μ ( x ) = exp Z x x p ( x ) dx . Above formulas are remarkably and explicitly applicable, which enable us to perform practical solving directly. Example 1 Find the solution of ty + 2 y = sin t, y ( π/ 2) = 1 , t > . First note that this equation is not separable. Divide the equation by t to obtain the standard form y + 2 t y = sin t t , which gives us p ( t ) = 2 /t and g ( t ) = (sin t ) /t. Thus, μ ( t ) = exp " Z t π/ 2 2 t dt # = 4 π 2 t 2 1 and y = π 2 4 t 2 " 1 + Z t π/ 2 4 π 2 t 2 sin t t dt # = π 2 4 t 2 " 1 + 4 π 2 (- t cos t + sin t ) t π/ 2 # = 1 t 2 π 2 4- 1- t cos t + sin t . 2 Example 2 Solve the initial value problem...
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