Solve the given differential equation by undetermined coefficients: $y''' + y'' = 7x^2$

Neetesh Kumar | November 05, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec05 (Homework) Question - 7
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Step-by-Step-Solution:

To solve this differential equation, we need to find both the complementary solution $y_c(x)$ and the particular solution $y_p(x)$.

Find the Complementary Solution, $y_c(x)$:

The complementary equation is:

$y''' + y'' = 0$

Rewrite this equation using the differential operator $D$:

$D^3 y + D^2 y = 0$

Factor out $y$:

$y(D^2)(D + 1) = 0$

This gives the characteristic equation:

$D^2(D + 1) = 0$

Solving for $D$, we get:

• $D = -1$, which gives the solution $y = C_1 e^{-x}$

• $D = 0$ (with multiplicity 2), which gives solutions $y = C_2$ and $y = C_3 x$

Therefore, the complementary solution is:

$y_c(x) = C_1 e^{-x} + C_2 + C_3 x$

Find the Particular Solution, $y_p(x)$:

Since the non-homogeneous term is $7x^2$, we assume a particular solution of the form:

$y_p(x) = Ax^4 + Bx^3 + Cx^2$

Differentiate $y_p(x)$ to find $y_p''$ and $y_p'''$:

• $y_p' = 4Ax^3 + 3Bx^2 + 2Cx$
• $y_p'' = 12Ax^2 + 6Bx + 2C$
• $y_p''' = 24Ax + 6B$ (since $y_p'''$ is derived from a polynomial of degree 2)

Substitute $y_p''$ and $y_p'''$ into the original differential equation:

$y''' + y'' = 7x^2$

Substituting in, we get:

$24Ax + 6B + 12Ax^2 + 6Bx + 2C = 7x^2$

$12Ax^2 + (24A + 6B)x + (6B+2C) = 7x^2$

Comparing terms on both sides:

$A = \frac{7}{12}, B = -\frac{7}{3}, C = \frac{7}{2}$

So we can write: $y_p = \frac{7}{12}x^4 - \frac{7}{3}x^3 + \frac{7}{2}x^2$

So, the $y = y_c(x) + y_p(x)$

Final Answer:

$y = \boxed{C_1 e^{-x} + C_2 + C_3 x + \frac{7}{12}x^4 - \frac{7}{3}x^3 + \frac{7}{2}x^2}$

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