# Thread: Second order differential equation

1. ## Second order differential equation

1. Find the general solution of each differential equation using the substitution given.
x(d^2 y)/(d x^2) – 2 dy/dx + x = 0; dy/dx = v

2. Use the substitution v = y^-2 to find the general solution of the differential equation
dy/dx + y/x = x^2 y^3
Find also y in terms of x, given that y = 1 at x = 1.

2. 1. Find the general solution of each differential equation using the substitution given.
x(d^2 y)/(d x^2) – 2 dy/dx + x = 0; dy/dx = v
what exactly don't you understand?, it's stright forward, you substitute and get a first order ODE.

2.

now use the chain rule:

$\displaystyle \begin{gathered} v = y^{ - 2} \hfill \\ v' = - 2y^{ - 3} y' \hfill \\ \end{gathered}$

thus:

$\displaystyle \begin{gathered} y = v^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} \hfill \\ y' = - \frac{1} {2}y^3 v' \hfill \\ \end{gathered}$

now substitute the resulting expressions in the DE:

$\displaystyle - \frac{1} {2}v^{ - {\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} v' + \frac{{v^{ - {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} }} {x} =$$\displaystyle x^2 v^{ - {\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}$

multiply the equationn by: $\displaystyle v^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}$
and you'll get a regular ODE...

3. Originally Posted by geton
1. Find the general solution of each differential equation using the substitution given.
x(d^2 y)/(d x^2) – 2 dy/dx + x = 0; dy/dx = v

2. Use the substitution v = y^-2 to find the general solution of the differential equation
dy/dx + y/x = x^2 y^3
Find also y in terms of x, given that y = 1 at x = 1.

1. Since $\displaystyle v = \frac{dy}{dx}$, it follows that $\displaystyle \frac{dv}{dx} = \frac{d^2 y}{dx^2}$.

Therefore $\displaystyle x \frac{dv}{dx} - 2 v + x = 0 \Rightarrow \frac{dv}{dx} - \frac{2}{x} v = -1 \,$.

Use integrating factor method to solve for v = v(x).

Then solve $\displaystyle \frac{dy}{dx} = v(x)$ for y = y(x).

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2. $\displaystyle v = \frac{1}{y^2} \Rightarrow \frac{dv}{dy} = -\frac{2}{y^3}$.

Using the chain rule: $\displaystyle \frac{dy}{dx} = \frac{dy}{dv} \times \frac{dv}{dx} = -\frac{y^3}{2} \times \frac{dv}{dx}$.

Sub into DE:

$\displaystyle -\frac{y^3}{2} \frac{dv}{dx} + \frac{y}{x} = x^2 y^3 \Rightarrow -\frac{y^2}{2} \frac{dv}{dx} + \frac{1}{x} = x^2 y^2 \Rightarrow - \frac{1}{2v} \frac{dv}{dx} + \frac{1}{x} = \frac{x^2}{v}$

$\displaystyle \Rightarrow \frac{dv}{dx} - \frac{2}{x} v = -2x^2$.

Solve for v = v(x) using the integrating factor method. Then $\displaystyle y^2 = \frac{1}{v(x)}$.

4. For question 1, it may need complementary function in this form Ae^mx + Be^mx.

Ax^3 + ½ x^2 + B.

How can I found complementary function by this equation?

5. Originally Posted by geton
For question 1, it may need complementary function in this form Ae^mx + Be^mx.

Ax^3 + ½ x^2 + B.

How can I found complementary function by this equation?
What's all this complementary function business?? As I've already said, to solve $\displaystyle \frac{dv}{dx} - \frac{2}{x} v = -1$, use the integrating factor method.

The integrating factor is $\displaystyle e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln |x|} = e^{\ln x^{-2}} = \frac{1}{x^2}$.

It follows that $\displaystyle v = x + C x^2$.

Then $\displaystyle \frac{dy}{dx} = x + C x^2 \Rightarrow y = \frac{1}{2} x^2 + \frac{C}{3} x^3 + B$.

Since $\displaystyle \frac{C}{3}$ is also an arbitrary constant, replace it with A, say.

Then $\displaystyle y = \frac{1}{2} x^2 + A x^3 + B$.