Math  /  Calculus

Questionb) (10 marks) Consider the following IVP problem: 14yy+10y=(2x3)2;y(0)=1;y(0)=0\frac{1}{4} y^{\prime \prime}-y^{\prime}+10 y=(2 x-3)^{2} ; y(0)=1 ; y^{\prime}(0)=0 i. Write the complementary equation for the given equation. ii. Write the auxiliary equation for the equation in part (i), solve it and hence obtain the general solution of the complementary equation. iii. Find a particular solution for the given equation. iv. Write the general solution for the given equation. v. Hence find the solution for the given IVP problem.

Studdy Solution
Use the initial conditions y(0)=1 y(0) = 1 and y(0)=0 y'(0) = 0 to find C1 C_1 and C2 C_2 .
First, apply y(0)=1 y(0) = 1 : 1=e0(C1cos(0)+C2sin(0))+15 1 = e^{0} (C_1 \cos(0) + C_2 \sin(0)) + \frac{1}{5} 1=C1+15 1 = C_1 + \frac{1}{5} C1=45 C_1 = \frac{4}{5}
Next, find y(x) y'(x) and apply y(0)=0 y'(0) = 0 : y(x)=ddx(e2x(C1cos(6x)+C2sin(6x)))+ddx(25x25825x+15) y'(x) = \frac{d}{dx} \left( e^{2x} (C_1 \cos(6x) + C_2 \sin(6x)) \right) + \frac{d}{dx} \left( \frac{2}{5}x^2 - \frac{58}{25}x + \frac{1}{5} \right)
y(0)=0Solve for C2 y'(0) = 0 \Rightarrow \text{Solve for } C_2
After computing derivatives and substituting: C2=130 C_2 = -\frac{1}{30}
Thus, the solution to the IVP is: y(x)=e2x(45cos(6x)130sin(6x))+25x25825x+15 y(x) = e^{2x} \left( \frac{4}{5} \cos(6x) - \frac{1}{30} \sin(6x) \right) + \frac{2}{5}x^2 - \frac{58}{25}x + \frac{1}{5}
The solution to the IVP is:
y(x)=e2x(45cos(6x)130sin(6x))+25x25825x+15 y(x) = e^{2x} \left( \frac{4}{5} \cos(6x) - \frac{1}{30} \sin(6x) \right) + \frac{2}{5}x^2 - \frac{58}{25}x + \frac{1}{5}

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