Thursday, 3 September 2015

`dy/dx = xcos(x^2)` Use integration to find a general solution to this differential equation.

To solve this differential equation, rewrite it as


`dy = xcos(x^2)dx`


Integrate both sides of the equation:


`y = int xcos(x^2)dx`


To take the integral on the right side of the equation, use the substitution method. Let `z(x) = x^2` .


Then, `dz = 2xdx` and integral becomes


`int xcos(x^2)dx = int(xdx) cos(x^2) = int 1/2 dz cosz = 1/2 int coszdz`


This is a simple trigonometric integral: `int cosz dz = sinz` . Substituting...

To solve this differential equation, rewrite it as


`dy = xcos(x^2)dx`


Integrate both sides of the equation:


`y = int xcos(x^2)dx`


To take the integral on the right side of the equation, use the substitution method. Let `z(x) = x^2` .


Then, `dz = 2xdx` and integral becomes


`int xcos(x^2)dx = int(xdx) cos(x^2) = int 1/2 dz cosz = 1/2 int coszdz`


This is a simple trigonometric integral: `int cosz dz = sinz` . Substituting the original variable, x, back into equation results in


`y = 1/2 sinx^2 + C` , where C is an arbitrary constant.


So, the general solution of the given differential equation is


`y = 1/2sinx^2 + C` .

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