Question

y'''+y''=8x^2

Answer 1

Answer on Question #58430 – Math – Differential Equations

Question

$$y''' + y'' = 8x^2.$$

Solution

The general solution will be the sum of the complementary solution and particular solution

$$y = y_{0+} y^*.$$

Find complementary solution $y''' + y'' = 0$.

Set up and solve the characteristic equation:

$$k^3 + k^2 = 0,$$ $$k_1 = k_2 = 0, \; k_3 = -1,$$

then

$$y_0 = C_1 + C_2 x + C_3 e^{-x}.$$

Find a particular solution of $y''' + y'' = 8x^2$ by the method of undetermined coefficients:

$$y^* = x^2 (Ax^2 + Bx + C) = Ax^4 + Bx^3 + Cx^2.$$

Solve for unknown constants $A, B, C$:

$$y^{*'} = 4Ax^3 + 3Bx^2 + 2Cx.$$ $$y^{*''} = 12Ax^2 + 6Bx + 2C.$$ $$y^{*'''} = 24Ax + 6B.$$

Substitute the particular solution into the differential equation:

$$24Ax + 6B + 12Ax^2 + 6Bx + 2C = 8x^2.$$ $$12Ax^2 + (24A + 6B)x + (6B + 2C) = 8x^2.$$

Solve the system:

$$\left\{ \begin{array}{l} 12A = 8, \\ 24A + 6B = 0, \Rightarrow \\ 6B + 2C = 0. \end{array} \right. \Rightarrow \left\{ \begin{array}{l} A = \frac{2}{3} \\ B = -4A = -\frac{8}{3} \\ C = -3B = 8 \end{array} \right.$$

Thus,

$$y ^ {*} = \frac {2}{3} x ^ {4} - \frac {8}{3} x ^ {3} + 8 x ^ {2},$$

then the general solution is given by

$$y = y _ {0} + y ^ {*} = C _ {1} + C _ {2} x + C _ {3} e ^ {- x} + \frac {2}{3} x ^ {4} - \frac {8}{3} x ^ {3} + 8 x ^ {2}.$$

Answer: $y = C_{1} + C_{2}x + C_{3}e^{-x} + \frac{2}{3} x^{4} - \frac{8}{3} x^{3} + 8x^{2}$ .

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