Solve the following differential equation.

{eq}\displaystyle \dfrac {d^2y}{dx^2} - 4 \dfrac {dy}{dx} + 3 y = 3 x^2 - 4 {/eq}


Solve the following differential equation.

{eq}\displaystyle \dfrac {d^2y}{dx^2} - 4 \dfrac {dy}{dx} + 3 y = 3 x^2 - 4 {/eq}

Complementary Solution of a Second Order Differential Equation:

Two functions are linearly dependent if one of them is obtained by multiplying the other function by a scalar, otherwise they are said to be two linearly independent functions. The complementary solution of a non-homogeneous differential equation of second order is the linear combination of two linearly independent functions.

Answer and Explanation: 1


$$\frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y=3x^2-4\\ $$

Calculate the solution of the homogeneous equation.

$$\begin{align} \frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y&=0 && \left[\textrm { Homogeneous equation } \right] \\[0.3cm] (D^2-4D+3)y& =0 && \left[ \textrm { Equation in terms of the differential operator D } \right] \\[0.3cm] \lambda^2 -4\lambda +3 & =0 && \left[ \textrm { Characteristic equation in terms of } \quad \lambda \right] \\[0.3cm] \end{align}\\ $$

  • Therefore, characteristic equation is: {eq}\,\, \lambda^2 -4\lambda +3 =0 {/eq}

Solve for {eq}\,\, \lambda \,\, {/eq} auxiliary equation.

$$\begin{align} \lambda^2 -4\lambda +3 & =0 && \left[ \textrm { Solve for } \lambda \right] \\[0.3cm] (\lambda-3) (\lambda-1) &= 0 && \left[ \textrm { Apply zero product property } \right] \\[0.3cm] \lambda _{1}& =1 \\[0.3cm] \lambda _{2}& =3 && \left[ \textrm { Different real roots } \right] \\[0.3cm] \end{align}\\ $$

  • Therefore, the functions of the homogeneous solution are: {eq}\,\, e^{x} \,\, {/eq} and {eq}\,\, e^{3x} \,\, {/eq}
  • Therefore, the complementary solution is: $$\,\, y_{h} (x) = C_1 e^{x} + C_2 e^{3x} $$

Find the particular solution, there are several methods for this calculation, we can use the method of undetermined coefficients. Find the particular solution.

$$\begin{align} \frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y&=3x^2-4\\[0.3cm] y_{p} (x) &=Ax^2+Bx+C &&\left[\text{ Solution proposed in the analyzed method } \right] \\[0.3cm] y'_{p}(x) &= 2Ax+B\\[0.3cm] y''_{p}(x) &=2A\\[0.3cm] \frac{d^2y}{dx^2}-4\frac{dy}{dx}+3y&=3x^2-4&&\left[ \text{ Substitute the derivatives } \right] \\[0.3cm] 2A-4(2Ax+B)+3(Ax^2+Bx+C)&=3x^2-4\\[0.3cm] 2A-8Ax-4B+3Ax^2+3Bx+3C&=3x^2-4\\[0.3cm] 3Ax^2+x(-8A+3B) +2A-4B+3C&=3x^2-4\\[0.3cm] 3Ax^2&=3x^2 \\[0.3cm] 3A&=3 &&\left[ \therefore A= 1 \right] \\[0.3cm] x(-8A+3B) &=0 &&\left[ \text{ Substitute: }A=1 \, \right] \\[0.3cm] -8(1)+3B &=0 \\[0.3cm] 3B&=8 &&\left[ \therefore B= \frac{8}{3} \right] \\[0.3cm] 2A-4B+3C &=-4 &&\left[ \text{ Substitute: }A=1 \, , \, B= \frac{8}{3} \, \right] \\[0.3cm] 2(1)-4\frac{8}{3} +3C &=-4\\[0.3cm] 3C &=\frac{14}{3} &&\left[ \therefore C= \frac{14}{9} \right] \\[0.3cm] \end{align}\\ $$

  • Therefore, the particular solution using the method of undetermined coefficients is: {eq}\,\, \bf{ y_{p} (x) = x^2+\frac{8}{3}x+\frac{14}{9} } {/eq}

  • Therefore, the general solution using the method of undetermined coefficients is: {eq}\,\, \bf{ y_{g} (x) = C_1 e^{x} + C_2 e^{3x}+ x^2+\frac{8}{3}x+\frac{14}{9} } {/eq}

Learn more about this topic:

Separable Differential Equation: Definition & Examples


Chapter 16 / Lesson 1

Discover what separable differential equations are and their uses. Learn to identify if an equation is separable and how to solve them through given examples.

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