Questions LLC
Login
or
Sign Up
Ask a New Question
Mathematics
Calculus
how do you solve the initial value problem by using separation of variables dy/dx=1/y^2, y(0)=4
1 answer
see other post. Please stop posting duplicate posts, I am getting the impression you are answer grazing.
You can
ask a new question
or
answer this question
.
Similar Questions
solve the initial value problem by separation of variables dy/dx=x2/y given y=-5 when x=3
Top answer:
follow the other post.
Read more.
how do you solve the initial value problem by using separation of variables dy/dx=-xy^2 when y(1)=-0.25
Top answer:
To solve the initial value problem dy/dx = -xy^2 with the initial condition y(1) = -0.25 using
Read more.
solve the initial value problem by separation of variables 8. dy/dx=x+1/xy, x>0, y(1)=-4
Top answer:
Please check that you have sufficient parentheses. The problem posted (implicitly) is dy/dx = x+
Read more.
solve the initial value problem by separation of variables dy/dx=y^(2)+1, y(1)=0
Top answer:
The integral of dy/(y^2 +1) equals the integral of dx. Integrate both sides from the y(1) = 0 point
Read more.
solve the initial value problem by separation of variables dy/dx=-x^2y^2, y(4)=4
Top answer:
These are all similar to your previous problem (under Billy):
Read more.
how do you solve the initial value problem by separation of variables dy/dx=-x^2y^2, y(4)=4
Top answer:
To solve the initial value problem dy/dx = -x^2y^2 with the initial condition y(4) = 4 using
Read more.
solve the initial value problem by separation of variables dy/dx=-xy^2, y(1)=-0.25
Top answer:
To solve the initial value problem dy/dx = -xy^2 with the initial condition y(1) = -0.25, we can use
Read more.
solve the initial value problem by separation of variables dy/dx=6x^2y and y(0)=4
Top answer:
dy/y = 6 x^2 dx ln y = 2 x^3 + c when x = 0, y = 4 ln 4 = c so ln y = 2 x^3 + ln 4 ln (y/4) = 2 x^3
Read more.
solve the initial value problem by separation of variables dy/dx=-xy^2, y(1)=-0.25
Top answer:
These are all similar to your previous problem (under Billy):
Read more.
solve the initial value problem by separation of variables dy/dx=-x^2y^2, y(4)=4
Top answer:
To solve the initial value problem dy/dx = -x^2y^2 with the initial condition y(4) = 4 using
Read more.
Related Questions
Using separation of variables technique, solve the following differential equation with initial condition dy/dx = (yx + 5x) /
Using the separation of variables technique, solve the following differential equation with initial condition:
(4x sqrt)(1 -
Using separation of variables, solve the following differential equation with initial conditions dy/dx = e^(2x+3y) and y(0) = 1.
Use separation of variables to find the solution to the differential equation:
4 (du/dt) = u^2, subject to the initial condition
Use separation of variables to find the solution to the differential equation:
4 (du/dt) = u^2, subject to the initial condition
Maximize P=60x+50y.
⎧ ⎪x+y≤80 ⎪ ⎪5x+10y≤560 ⎪ ⎪50x+20y≤1600 ⎪ ⎪x≥0 ⎨ ⎪y≥0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
how do you solve the initial value problem by using separation of variables dy/dx=1+x/(sqrt of y), y(2)=9
Solve the differential equation y'=3t^2+4. Solve the initial value problem y(0)=3.
Separation of variables! My work: dy/dt=
How would I use separation of variables to solve the initial value problem?
dy/dx = (y +5)(x+2) I multiply both sides by dx so:
solving the initial value problem by separation of variables dy/dx=(4(sqrt of y)lnx))/x, y(e)=9