How do you integrate #ydx = 2(x+y)dy#?
By signing up, you agree to our Terms of Service and Privacy Policy
To solve the given differential equation ( y , dx = 2(x+y) , dy ), we'll first rearrange terms and then integrate both sides. Rearranging, we get:
[ \frac{1}{x} , dx = \frac{2}{y} , dy ]
Integrating both sides, we have:
[ \int \frac{1}{x} , dx = \int \frac{2}{y} , dy ]
[ \ln|x| = 2 \ln|y| + C ]
Where ( C ) is the constant of integration.
Using logarithmic properties, we can simplify:
[ \ln|x| = \ln|y|^2 + C ]
[ \ln|x| = \ln|y|^2 + \ln|e^C| ]
[ \ln|x| = \ln|y|^2 \cdot |e^C| ]
[ \ln|x| = \ln(C \cdot |y|^2) ]
[ |x| = C \cdot |y|^2 ]
[ x = Cy^2 ]
Where ( C ) is an arbitrary constant. This is the solution to the differential equation.
By signing up, you agree to our Terms of Service and Privacy Policy
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.
- Find the arc length of the function #y=1/2(e^x+e^-x)# with parameters #0\lex\le2#?
- How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=x^2# and #y^2=x# rotated about the x-axis?
- What is the surface area produced by rotating #f(x)=1/(x+1), x in [0,3]# around the x-axis?
- How do you find the volume of the solid generated by revolving the plane region bounded by the graphs of #x^2 = y - 2# and 2y - x -2 = 0 about the line y =3 with x =0, x=1?
- Consider a particle moving along the x-axis where x(t) is the position of the particle at time t A particle, initially at rest, moves along the x-axis such that its acceleration at time t > 0 is given by a(t)=5cos(t). At t=0, its position is x=2?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7