Solving the Differential Equation ( t^2x 2tx - 2x 0 ) with Initial Value Problem

Consider the differential equation:

( t^2x 2tx - 2x 0 )

with the assumption that the solution has the form ( x(t) t^r ), where ( r ) is a constant to be determined. This method is often referred to as the method of substitution.

Determining the General Solution

First, we compute the first and second derivatives of ( x(t) t^r ):

First derivative: ( x(t) r t^{r-1} ) Second derivative: ( x'(t) r(r-1) t^{r-2} )

Substitute ( x(t) ) and its derivatives into the differential equation:

( t^2 cdot r(r-1) t^{r-2} 2t cdot r t^{r-1} - 2 t^r 0 )

Simplify each term:

First term becomes: ( r(r-1) t^r ) Second term becomes: ( 2r t^r ) Third term is: ( -2 t^r )

Combine these terms:

( r(r-1) t^r 2r t^r - 2 t^r 0 )

Factoring out ( t^r ) (since ( t^r eq 0 ) for ( t eq 0 )), we have:

( t^r [r(r-1) 2r - 2] 0 )

This simplifies to:

( r^2 - r - 2 0 )

This is a quadratic equation, which can be factored as:

( (r - 1)(r 2) 0 )

Solving for ( r ), we get:

( r 1 ) and ( r -2 )

The general solution of the differential equation can be expressed as a linear combination of the two independent solutions:

( x(t) C_1 t^1 C_2 t^{-2} C_1 t - frac{C_2}{t^2} )

where ( C_1 ) and ( C_2 ) are arbitrary constants.

Initial Value Problem (IVP)

To specify the initial conditions for an IVP, we need to define the values of ( x(t) ) and ( x'(t) ) at a specific point ( t_0 ), e.g., ( t_0 1 ).

Assume:

( x(1) x_0 ) and ( x'(1) v_0 )

We need to compute ( x(t) ) and ( x'(t) ):

From ( x'(t) C_1 - 2 cdot frac{C_2}{t^3} ), substitute into the initial conditions: ( x(1) C_1 - C_2 x_0 ) ( x'(1) C_1 - 2C_2 v_0 )

We now have a system of equations:

( C_1 - C_2 x_0 ) ( C_1 - 2C_2 v_0 )

Solving this system to find ( C_1 ) and ( C_2 ):

From equation 1, we can express ( C_2 ) in terms of ( C_1 ):

( C_2 x_0 - C_1 )

Substituting into equation 2:

( C_1 - 2(x_0 - C_1) v_0 )

This simplifies to:

( C_1 - 2x_0 2C_1 v_0 )

Combining like terms:

( 3C_1 - 2x_0 v_0 )

Solving for ( C_1 ):

( C_1 frac{v_0 2x_0}{3} )

Substituting ( C_1 ) back into equation 1 to find ( C_2 ):

( C_2 x_0 - frac{v_0 2x_0}{3} frac{3x_0 - (v_0 2x_0)}{3} frac{x_0 - v_0}{3} )

Thus, the general solution of the IVP is:

( x(t) frac{v_0 2x_0}{3} t - frac{x_0 - v_0}{3t^2} )

This expression gives the solution of the differential equation with the specified initial conditions.