Solving differential equations

[hahaha158]

Homework Statement

solve

(y^15)x(dy/dx)=1+x

using the initial condition y(1)=3

express y^16 in terms of x

Homework Equations

The Attempt at a Solution

i change the equation to the form

y^15 dy = (1+x)/x dx
integrating both sides i got

y^16/16=x+lnx+C

To solve for C i changed this to

y=(16(x+lnx))^(1/16)+C

plugging in y=3 and x=1 i get

3=1.189+C

so C=1.81079

So my answer was y^16= 16(x+lnx)+1.81079

this is incorrect, can anyone please explain what i may be doing wrong?

 

[ehild]

Homework Statement

solve

(y^15)x(dy/dx)=1+x

using the initial condition y(1)=3

express y^16 in terms of x

Homework Equations

The Attempt at a Solution

i change the equation to the form

y^15 dy = (1+x)/x dx
integrating both sides i got

y^16/16=x+lnx+C

Correct so far, except you shoiuld write ln(|x|).

To solve for C i changed this to

y=(16(x+lnx))^(1/16)+C

that is wrong. It should be y=[16(x+ln(x)+C)]^1/16,
but it has no sense. Just substitute x=1 and y=3 in the original equation
y^16/16=x+lnx+C and isolate C.

ehild

 

[hahaha158]

Correct so far, except you shoiuld write ln(|x|).



that is wrong. It should be y=[16(x+ln(x)+C)]^1/16,
but it has no sense. Just substitute x=1 and y=3 in the original equation
y^16/16=x+lnx+C and isolate C.

ehild

thank you

 

[HallsofIvy]

Homework Statement

solve

(y^15)x(dy/dx)=1+x

using the initial condition y(1)=3

express y^16 in terms of x

Homework Equations

The Attempt at a Solution

i change the equation to the form

y^15 dy = (1+x)/x dx
integrating both sides i got

y^16/16=x+lnx+C

To solve for C i changed this to

y=(16(x+lnx))^(1/16)+C

I see know reason to solve for y (especially since you were asked to "express y^16 in terms of x"). From
y^16/16= x+ ln|x|+ C, setting x= 1, y= 3 gives
3^16/16= 2690420.0625= 1+ C so that C= 2690419.0625

plugging in y=3 and x=1 i get

3=1.189+C

so C=1.81079

So my answer was y^16= 16(x+lnx)+1.81079

this is incorrect, can anyone please explain what i may be doing wrong?