Using the method introduced in Question 26, find the solutions of the following initial-value problems: Data from

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Using the method introduced in Question 26, find the solutions of the following initial-value problems:

dx (a) x- = dt d.x dt 2 x(0) = 4, dx dt -(0)=1

(b) dx dt (d) dx dt d.x dx e, x(1)=1,(1) = 0 dt dt (c) = xdx dt dx 1 / dx + dt 2 dt x(0) = 2, d.x -(0) dt =


Data from Question 26

Show that by making the substitution

d.x dt and noting that dx du dt dt the equation dx dt may be expressed as v= du dx d.x -X = || ax=a dt du dx

Show that the solution of this equation is v = 1/2x2 + C and hence find x(t). This technique is a standard method for solving second-order differential equations in which the independent variable does not appear explicitly. Apply the same method to obtain the solutions of the differential equations

(a) (b) (2) dx TIP X-P dt X-P ZIP || || xp -d dt Xp dt = (dr) (x - - ) |2x

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