(a) By inspection, find a particular solution of $y'' + 2y = 6$. (b) By inspection, find a particular solution of $y'' + 2y = -12x$. (c) Find a particular solution of $y'' + 2y = -12x + 6$. (d) Find a particular solution of $y'' + 2y = 24x + 3$.

Neetesh Kumar | November 06, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec01 (Homework) Question - 7
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Step-by-Step-Solution:

(a) By inspection, find a particular solution of $y'' + 2y = 6$.

Since the right side is a constant ($6$), we can assume a particular solution of the form: $y_p(x) = A$

Substitute $y_p(x) = A$ into the differential equation: $y_p'' + 2y_p = 6$

Since $y_p'' = 0$, this simplifies to: $2A = 6$

Solve for $A$: $A = 3$

Thus, a particular solution is: $y_p(x) = \boxed{3}$

(b) By inspection, find a particular solution of $y'' + 2y = -12x$.

Since the right side is a linear term ($-12x$), we can assume a particular solution of the form: $y_p(x) = Bx$

Substitute $y_p(x) = Bx$ into the differential equation: $y_p'' + 2y_p = -12x$

Since $y_p'' = 0$, this simplifies to: $2(Bx) = -12x$

Comparing coefficients, we get: $2B = -12$

Solve for $B$: $B = -6$

Thus, a particular solution is: $y_p(x) = \boxed{-6x}$

(c) Find a particular solution of $y'' + 2y = -12x + 6$.

Since the right side is $-12x + 6$, which is a linear polynomial, we can assume a particular solution of the form: $y_p(x) = Cx + D$

Substitute $y_p(x) = Cx + D$ into the differential equation: $y_p'' + 2y_p = -12x + 6$

Since $y_p'' = 0$, this simplifies to: $2(Cx + D) = -12x + 6$

Expanding and matching coefficients: $2Cx = -12x \quad \Rightarrow \quad C = -6$ $2D = 6 \quad \Rightarrow \quad D = 3$

Thus, a particular solution is: $y_p(x) = \boxed{-6x + 3}$

(d) Find a particular solution of $y'' + 2y = 24x + 3$.

Since the right side is $24x + 3$, we can assume a particular solution of the form: $y_p(x) = Ex + F$

Substitute $y_p(x) = Ex + F$ into the differential equation: $y_p'' + 2y_p = 24x + 3$

Since $y_p'' = 0$, this simplifies to: $2(Ex + F) = 24x + 3$

Expanding and matching coefficients:

$2E = 24 \quad \Rightarrow \quad E = 12$

$2F = 3 \quad \Rightarrow \quad F = \frac{3}{2}$

Thus, a particular solution is: $y_p(x) = \boxed{12x + \frac{3}{2}}$

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