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W6 Second Order Differential Equations

Form

  1. What is the form of 2nd ODE?
    • F(x,y,dydx,d2ydx2)F(x,y,\frac{dy}{dx},\frac{d^2y}{dx^2})
  2. What is the form of a Linear 2nd ODE?
    • d2ydx2+P(x)dydx+Q(x)y=R(x)\frac{d^2y}{dx^2}+P(x)\frac{dy}{dx}+Q(x)y=R(x)
  3. When R(x)=0R(x)=0, the ODE is…
    • Homeogeneous, (H)
  4. When R(x)0R(x)\ne0, the ODE is…
    • Inhomeogeneous, (IH)

Homeogenous, Linear 2nd ODE

  1. What is the general form of a homeogenous, linear 2nd ODE?
    • ay+by+cy=0ay''+by'+cy=0
  2. What is the general form of the general solution of a homeogenous, linear 2nd ODE?
    • GS(H)=c1y1(x)+c2y2(x)GS(H)=c_1y_1(x)+c_2y_2(x)
  3. How do you find the general solution of a homeogenous, linear 2nd ODE?
    • Step 1: Write the differential equation in the form ay+by+cy=0ay''+by'+cy=0
    • Step 2: Find the corresponding characteristic equation aλ2+bλ+c=0a\lambda^2+b\lambda+c=0
      • Try y(x)=eλxy(x)=e^{\lambda x}
      •     y(x)=λeλx\implies y'(x)=\lambda e^{\lambda x}
      •     y(x)=λ2eλx\implies y''(x)=\lambda^2 e^{\lambda x}
      • (aλ2+bλ+c=0)eλx=0(a\lambda^2+b\lambda+c=0)e^{\lambda x}=0
        • eλx0e^{\lambda x}\ne0
      • aλ2+bλ+c=0a\lambda^2+b\lambda+c=0 (Characteristic Equation)
      • λ=bb24ac2a\lambda=\frac{-b\sqrt{b^2-4ac}}{2a} (Quadratic Equation)
    • Step 3: Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or coplex conjugate roots.
    • Case 1: b24ac>0b^2-4ac>0
      • 2 distinct values: λ1,λ2\lambda_1, \lambda_2
      • 2 linearly independent solution:
        • y1(x)=eλ1xy_1(x)=e^{\lambda_1x}
        • y2(x)=eλ2xy_2(x)=e^{\lambda_2x}
      • General Solution:
        • y(x)=Aeλ1x+Beλ2xy(x)=Ae^{\lambda_1x}+Be^{\lambda_2x}, where A,BRA, B\in\mathbb{R} constant and for all xRx\in\mathbb{R}
    • Case 2: b24ac=0b^2-4ac=0
      • 1 real value: lambda=b2alambda=-\frac{b}{2a}
      • solutions:
        • eλxe^{\lambda x}
        • xeλxxe^{\lambda x}
      • General Solution
        • y(x)=Aeλx+Bxeλx=(A+Bx)eλxy(x)=Ae^{\lambda x}+Bxe^{\lambda x}=(A+Bx)e^{\lambda x}, where A,BRA, B\in\mathbb{R} constants and for all xRx\in\mathbb{R}
    • Case 3: b24ac<0b^2-4ac<0
      • 2 complex conjugate values:
        • λ1=α+iβ\lambda_1=\alpha +i\beta
        • λ2=αiβ\lambda_2=\alpha -i\beta
      • 2 complex linearly independent solutions
        • y1(x)=e(α+iβ)xy_1(x)=e^{(\alpha+i\beta)x}
        • y2(x)=e(αiβ)xy_2(x)=e^{(\alpha-i\beta)x}
      • 2 real linearly independent solutions
        • y1(x)=eαxcos(βx)y_1(x)=e^{\alpha x}\cos(\beta x)
        • y2(x)=eαxsin(βx)y_2(x)=e^{\alpha x}\sin(\beta x)
      • General Solution:
        • y(x)=Aeαxcos(βx)+Beαxsin(βx)y(x)=Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x), where A,BRA, B\in\mathbb{R} constants and for all xRx\in\mathbb{R}