Question

Solve the DE (x^2)y' + xy = (x^3)(y^2) , subject to y(0) = -0.15

Solve the DE (x^2)y' + xy = (x^3)(y^2) , subject to y(0) = -0.15

Math   Advanced-Math   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Solve the IVP: , y(0)=3. 2) Solve the DE: . y' = xy^2/ (x^2 +1)
Solve the IVP: , y(0)=3. 2) Solve the DE: . y' = xy^2/ (x^2 +1)
solve differential equation ((x)2 - xy +(y)2)dx - xydy = 0 solve differential equation (x^2-xy+y^2)dx -...
solve differential equation ((x)2 - xy +(y)2)dx - xydy = 0 solve differential equation (x^2-xy+y^2)dx - xydy = 0
Solve the DE: x^2y''-(x^2+2x)y^2+(x+2)y=0, x > 0 if you know that y=x is a solution of...
Solve the DE: x^2y''-(x^2+2x)y^2+(x+2)y=0, x > 0 if you know that y=x is a solution of this equation.
Solve the differential equation y^' − xy = e^x   y(0) = 2
Solve the differential equation y^' − xy = e^x   y(0) = 2
Solve the Homogeneous differential equation (7 y^2 + 1 xy)dx - 1 x^2 dy = 0...
Solve the Homogeneous differential equation (7 y^2 + 1 xy)dx - 1 x^2 dy = 0 (a) A one-parameter family of solution of the equation is y(x) = (b) The particular solution of the equation subject to the initial condition y(1) =1/7.
Solve the DE: x(y+1)dx + [(y^2+1)csc(3x)]dy=0
Solve the DE: x(y+1)dx + [(y^2+1)csc(3x)]dy=0
(3) solve the homogeneous equation x^2dydx= x^2+ xy+ y^2
(3) solve the homogeneous equation x^2dydx= x^2+ xy+ y^2
Solve the equation. (2x^3+xy)dx+(x^3y^3-x^2)dy=0 give answer in form F(x,y)=c
Solve the equation. (2x^3+xy)dx+(x^3y^3-x^2)dy=0 give answer in form F(x,y)=c
solve by separation variable 3, xydx-(1+x^2) dy=0 4, xy'+y=y^2 5,yy'=xe^2+y^2 6, xsiny.y'=cosy 7, y'=x^2 y/1+x^3 xsiny.y'+cosy
solve by separation variable 3, xydx-(1+x^2) dy=0 4, xy'+y=y^2 5,yy'=xe^2+y^2 6, xsiny.y'=cosy 7, y'=x^2 y/1+x^3 xsiny.y'+cosy
use variation of parameter solve DE y"+y'-y=x initial conditions y(0)=0 y'(0)=0
use variation of parameter solve DE y"+y'-y=x initial conditions y(0)=0 y'(0)=0
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT