How do you use the differential equation #dy/dx=(2x)/sqrt(2x^2-1)# to find the equation of the function given point (5,4)?

1 Answer
Noah G
Mar 4, 2017

The solution is #y = sqrt(2x^2 - 1) - 3#.

Explanation:

This is a separable differential equation.

#dy = (2x)/sqrt(2x^2 - 1) dx#

Integrate both sides.

#int dy = int (2x)/sqrt(2x^2 - 1) dx#

It's true that trig substitution could be used to solve this integral, but a substitution would be easier.

#:.# Let #u = 2x^2 - 1#. Then #du = 4x dx# and #dx = (du)/(4x)#.

#int dy = int (2x)/sqrt(u) * (du)/(4x)#

#int dy = 1/2int 1/sqrt(u)#

#int dy = 1/2int u^(-1/2)#

#y = 1/2(2u^(1/2)) + C#

#y = u^(1/2) + C#

#y = (2x^2 - 1)^(1/2) + C#

We now solve for #C# using the information given. We know that when #x = 5#, #y = 4#, so we can say the following:

#4 = sqrt(2(5)^2 - 1) + C#

#4 = sqrt(49) + C#

#4 - 7 = C#

#C = -3#

The solution is therefore #y = sqrt(2x^2 - 1) - 3#.

Hopefully this helps!

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