How do you find the particular solution to #(du)/(dv)=uvsinv^2# that satisfies u(0)=1?

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Answer 1 · 2017-01-11T14:42:57

#u(v) = e^(sin^2(v^2/2))#

We can find the general solution by separating the variables:

#(du)/(dv) = uv sin (v^2)#

#(du)/u = vsin(v^2) dv#

#int (du)/u = int vsin(v^2) dv#

#ln u = 1/2int sin(v^2) d(v^2) = -1/2 cos(v^2) +C#

To determine #C# we note that for #v=0# we have:

#u(0) = 1 => ln(u(0)) = 0#

so

#-1/2 cos (0) + C = 0#

#-1/2 +C = 0#

#C=1/2#

The particular solution we look for is then:

#ln u = -1/2 cos(v^2) +1/2 = 1/2(1-cos(v^2)) = sin^2(v^2/2)#

#u(v) = e^(sin^2(v^2/2))#