Understanding Engineering Mathematics

Problem 15.6

Find the general solution of the DEs:

(i)

dy

dx

=xcosx (ii)

dy

dx

=cos^2 y

In (ii) find the particular solution satisfyingy=

p

4

whenx=0.

Check your answers by substituting back into the equations.

(i) Direct integration gives

y=

∫

xcosxdx+C

and the task is simply to do the integration. Integration by parts (273

➤

)gives

y=xsinx−

∫

sinxdx+C

=xsinx+cosx+C

which is the required solution, as you should check by substituting back into the DE.

(ii)

dy

dx

=cos^2 y

In this case, turn both sides upside down

dx

dy

=

1

cos^2 y

=sec^2 y

Integrating:

x=

∫

sec^2 ydy+C=tany+C

This is the implicit form of the solution (91

➤

). For the explicit form we solve fory:

tany=x−C

So, as you can check in the DE, the general solution is

y=tan−^1 (x−C)

Ify=

π

4

whenx=0thenwehave

y=

π

4

=tan−^1 (−C)

from whichC=−1 and the required solution is

y=tan−^1 (x+ 1 )