How to you find the general solution of #dy/dx=xcosx^2#?

Answer 1

First we notice that

#d(sinx^2)/dx=2xcosx^2#

or

#xcosx^2=1/2*(dsinx^2/dx)#

Hence the problem becomes

#dy/dx=1/2*(dsinx^2)/dx#
Integrate both sides with respect to #x# and we have
#int dy/dx*dx =1/2*int (dsinx^2/dx)dx#
#y=1/2*sinx^2+c#

The general solution is

#y(x)=1/2*sinx^2+c#
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Answer 2

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

#dy/dx = xcosx^2#
#dy = xcosx^2dx#
#int dy = int xcosx^2 dx#
THIS SOLUTION IS ONLY CORRECT IF THE PROBLEM IS WRITTEN CORRECTLY. The solution would be different if the problem is #dy/dx = xcos^2x#.
Let #u = x^2#. Then #du = 2xdx# and #dx = (du)/(2x)#.
#intdy = intxcosu (du)/(2x)#
#intdy = int 1/2cosudu#
#y = 1/2sinu + C#
#y = 1/2sin(x^2) + C#

Hopefully this helps!

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Answer 3

To find the general solution of the differential equation ( \frac{{dy}}{{dx}} = x\cos(x^2) ), you can separate variables and integrate both sides with respect to ( x ). The steps are as follows:

  1. Separate variables: [ \frac{{dy}}{{\cos(x^2)}} = x , dx ]

  2. Integrate both sides: [ \int \frac{{dy}}{{\cos(x^2)}} = \int x , dx ]

  3. Integrate the left side using a substitution. Let ( u = x^2 ), then ( du = 2x , dx ). [ \frac{1}{2} \int \frac{{dy}}{{\cos(u)}} , du = \frac{1}{2} \int x , dx ]

  4. The integral ( \int \frac{{dy}}{{\cos(u)}} , du ) can be evaluated using a trigonometric substitution or known trigonometric identities.

  5. Once you integrate both sides, you'll have the solution in terms of ( y ) and ( u ), which is ( x^2 ). You can then replace ( u ) with ( x^2 ) to obtain the solution in terms of ( x ).

This process will yield the general solution of the given differential equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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