Home page for accesible maths 6 Chapter 6 contents

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

6.6 Further example

Example.

Find the general solution to the equation d⁢xd⁢t=t⁢sin⁡a⁢t\frac{dx}{dt}=t\sin at, where aa is a constant. Find the particular solution with x⁢(0)=1x(0)=1.

Solution. We integrate both sides to obtain

x⁢(t)=∫t⁢sin⁡a⁢t⁢d⁢t=-ta⁢cos⁡a⁢t+1a⁢∫cos⁡a⁢t⁢d⁢tx(t)={\int t\sin at\,dt=\frac{-t}{a}\cos at+\frac{1}{a}\int\cos at\,dt}
=-ta⁢cos⁡a⁢t+1a2⁢sin⁡a⁢t+c={\frac{-t}{a}\cos at+\frac{1}{a^{2}}\sin at+c}

which is the general solution. For the particular solution, we require x⁢(0)=1x(0)=1 and therefore -0+0+c=1,{-0+0+c=1,} whence

x⁢(t)=1-ta⁢cos⁡a⁢t+1a2⁢sin⁡a⁢t.x(t)={1-\frac{t}{a}\cos at+\frac{1}{a^{2}}\sin at.}