Home > Calculus > Solving Separable Differential Equations

How do you find the general solution to #dy/dx=2yx+yx^2#?

Answer 1

Cameron Alexander

#y = Ce^(x^2 + x^3 /3 )#

#dy/dx=2yx+yx^2#

#dy/dx=y(2x+x^2)#

#1/y dy/dx=2x+x^2#

#int \ 1/y dy/dx \ dx=int \ 2x+x^2 \ dx#

#ln y = x^2 + x^3 /3 + C#

#y = e^(x^2 + x^3 /3 + C)#

#y = Ce^(x^2 + x^3 /3 )#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isabella Ackerman

To find the general solution to the differential equation ( \frac{dy}{dx} = 2yx + yx^2 ), we can use separation of variables. First, we'll rewrite the equation as:

[ \frac{dy}{dx} = y(2x + x^2) ]

Next, we'll separate the variables by dividing both sides by ( y(2x + x^2) ):

[ \frac{1}{y} , dy = (2x + x^2) , dx ]

Now, we integrate both sides:

[ \int \frac{1}{y} , dy = \int (2x + x^2) , dx ]

[ \ln|y| = x^2 + x^3/3 + C ]

Where ( C ) is the constant of integration. Finally, we'll exponentiate both sides to solve for ( y ):

[ y = e^{x^2 + x^3/3 + C} ]

This is the general solution to the differential equation ( \frac{dy}{dx} = 2yx + yx^2 ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.