Match the following differential equations with their solutions. The symbols A, B, C in the solutions stand for arbitrary constants. You must get all of the answers correct to receive credit.
1. d^2y/dx^2 + 25y = 0
2. dy/dx = 2xy/x^2 - 5y^2
3. d62y/dx^2 + 16 dy/dx + 64y = 0
4. dy/dx = 10xy
5. dy/dx + 24x^2y = 24 x^2
A. y = Ce^-8x^3 + 1
B. 3yx^2 - 5y^3 = C
C. y = Ae^-8x + Bxe^-8x
D. y = Ae^5x^2
E. y = A cos(5x) + B sin(5x)

Respuesta :

Answer:

[tex]1 \rightarrow E, 2\rightarrow B, 3\rightarrow C, 4\rightarrow D, 5\rightarrow A[/tex]

Step-by-step explanation:

1.     [tex]\frac{d^2y}{dx^2}+25y=0[/tex]

    The characteristic equation for the given differential equation is:

        [tex]r^{2} +25=0[/tex]

    [tex]\Rightarrow r^2=-25[/tex]

    [tex]\Rightarrow r=\pm 5i[/tex]

   Since the roots are complex

Now, the general solution is:

 [tex]y=A\cos 5x+B\sin 5x[/tex]

2.    [tex]\frac{dy}{dx}=\frac {2xy}{x^2}-5y^2[/tex]

     [tex]\Rightarrow \frac{dy}{dx}-\frac 2xy=-5y^2[/tex]

  Divide both sides by [tex]y^{-1}[/tex]

  Let, [tex]v=y^{-1} \Rightarrow \frac{dv}{dx}=-y^{-2}\frac{dy}{dx}[/tex]

    [tex]\Rightarrow -\frac{dv}{dx}-\frac 2xv=-5[/tex]

    [tex]\Rightarrow \frac{dv}{dx}+\frac 2xv=5[/tex]

 Here, [tex]p(x)=\frac 2x\; \text{and}\;\; q(x)=5[/tex]

 I.F. [tex]=e^{\int \frac 2xdx}=x^2[/tex]

 Now, the general solution is:

    [tex]vx^2=\int x^2 5dx=\frac {5x^3}3+c[/tex]

      [tex]\Rightarrow \frac {x^2}y-\frac {5x^3}3=c[/tex]

      [tex]\Rightarrow 3x^2-5x^3y=Cy[/tex]

3.     [tex]\frac{d^2y}{dx^2}+16\frac{dy}{dx}+64y=0[/tex]

   The characteristic equation is:

     [tex]r^2+16r+64=0[/tex]

  [tex]\Rightarrow r^2+8r+8r+64=0[/tex]

  [tex]\Rightarrow r(r+8)+8(r+8)=0[/tex]

  [tex]\Rightarrow (r+8)(r+8)=0[/tex]

  [tex]\Rightarrow r=-8,-8[/tex]

 Since the roots are real and repeated.

 Now, the general solution is:

 [tex]y=Ae^{-8x}+Bxe^{-8x}[/tex]

4.    [tex]\frac {dy}{dx}=10xy[/tex]

   [tex]\Rightarrow \frac {dy}{y}=10xdx[/tex]

 Integrating both sides

     [tex]\int\frac {dy}y=\int 10xdx+\log c[/tex]

   [tex]\Rightarrow \log y=5x^2+\log c[/tex]

   [tex]\Rightarrow y=e^{5x^2}+c[/tex]

5.      [tex]\frac {dy}{dx}+24x^2y=24x^2[/tex]

      Here, [tex]p(x)=24x^2 \; \text{and}\;\; q(x)=24x^2[/tex]

  I.F.[tex]= e^{\int 24x^2dx}=e^{8x^3}[/tex]

  Now, the general solution is:

     [tex]y.e^{8x^3}=\int 24x^2 e^{8x^3}dx=24\int x^2e^{8x^3}dx[/tex]

               Let, [tex]8x^3=t \Rightarrow 24x^2dx=dt\Rightarrow x^2dx=\frac {dt}{24}[/tex]

   [tex]\Rightarrow ye^{8x^3}=\int e^tdt[/tex]

   [tex]\Rightarrow ye^{8x^3}=e^{8x^3}+c[/tex]

   [tex]\Rightarrow y=1+ce^{-8x^3}[/tex]

 

     

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