Working with Differential Equations
Hi,
For this problem please state the method you used and show the work required to obtain the answer.
Find the complete solution to this equation problem.
y^3 - 2y^2 + y^1 = 2- 24e^(x) + 40e^(5x)
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SOLUTION This solution is FREE courtesy of BrainMass!
<br>y^3 - 2y^2 + y^1 = 2- 24e^(x) + 40e^(5x) ....(1)
<br>
<br>
<br>First set the right side equal to 0 and find the solution as in the
<br>case of a homogeneous equation (this is called complementary function
<br>y1) and then by trial find a particular solution (particular integral
<br>y2)
<br>
<br>Now the general solution is given by,
<br>
<br> Y = y1 + y2
<br>
<br>First evaluate y1
<br>................
<br>
<br>y^3 - 2y^2 + y^1 = 0
<br>
<br>ie, y (y^2 - 2y +1) = 0 This gives y = 0, +1, +1
<br>
<br>Now our y1 is given by
<br>
<br>y1 = C1 e^(0 x) + (C2 x+C3) e^(1 x)
<br>
<br>= C1 + (C2 x + C3) e^x ........(2)
<br>
<br>
<br>Now Find y2
<br>...........
<br>
<br>We will use the method of undetermined coefficients to find y2,
<br>
<br>Let y2 = Ko + C1 e^x + C2 e^5x ......(3)
<br>
<br>(If, for example, a term 4x were present in the original equation, we
<br>will have a K1 x term added with this and so on for higher powers of x)
<br>
<br>We will determine Ko, C1 and C2 in this by the above said method
<br>For this find the derivatives y2^3, y2^2, y2^1 and substitute them in
<br>our original equation.
<br>
<br>y2^1 = 0 + C1 e^x + 5 C2 e^5x
<br>
<br>y2^2 = C1 e^x + 25 C2 e^5x
<br>
<br>y2^3 = C1 e^x + 125 C2 e^5x
<br>
<br>Substitute back in (1)
<br>
<br>(C1 e^x + 125 C2 e^5x) -2(C1 e^x + 25 C2 e^5x) + C1 e^x + 5 C2 e^5x
<br>
<br> = 2- 24e^(x) + 40e^(5x)
<br>
<br>Simplify the left side and equate the coefficients of like terms and
<br>find C1, K0 and C2
<br>
<br>We get, Ko = 0, C1 = 0, C2 = 40/80 = 1/2
<br>
<br>Give these values to (3) to get our original y2
<br>
<br>y2 = 0 + 0 + (1/2) e^(5x)
<br>
<br>Now the general solution of the given differential equation is,
<br>
<br>Y = y1 + y2
<br>
<br> = C1 + (C2 x + C3) e^x + (1/2) e^(5x) Answer
<br>
<br>
<br>
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