Question

Solve the initial-value problem y'+x(y^2+y)=0, y(2)=1.

Answer 1

Here's my work:

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

dy/(y^2+y)=-x*dx

A/y+B/(y+1)=1

A=1, B=-1

y/(y+1)=ce^(-x^2/2)

How do I solve for y?

Answer 2

\[\ln y-\ln \left( y+1 \right)=-\frac{ x^2 }{ 2 }+c,\]
when x=2,y=1
\[\ln 1-\ln 2=-\frac{ 4 }{ 2 }+c,0-\ln 2+2=c,c=2-\ln 2\]
\[\ln \frac{ y }{ y+1 }=-\frac{ x^2 }{ 2 }+2-\ln 2\]
\[\ln 2+\ln \frac{ y }{ y+1 }=\frac{ 4-x^2 }{ 2 }\]
\[2 \ln \frac{ 2y }{ y+1 }=4-x^2\]
\[\left( \frac{ 2y }{ y+1 } \right)^2=e ^{4-x^2}\]

Answer 3

@amistre64 @ganeshie8

Answer 4

@satellite73

Answer 5

How do I solve for y from there? Please check my work first.

Answer 6

\[\frac{ y }{ y+1 }=\frac{ y+1-1 }{ y+1 }=1-\frac{ 1 }{ y+1 }=ce ^{\frac{ -x^2 }{ 2 }}\]
\[\frac{ 1 }{ y+1 }=1-c e ^{\frac{ -x^2 }{ 2 }},y+1=\frac{ 1 }{ 1-c ^{\frac{ -x^2 }{ 2 }} }\]
y=?

Answer 7

Wait a minute. Let me work it out.

Answer 8

So y+1=1-e^(x^2/2)/c, right?

Answer 9

\[y=\frac{ 1-1+c e ^{\frac{ -x^2 }{ 2 }} }{ 1-c e ^{\frac{ -x^2 }{ 2 }} }=\frac{ c e ^{\frac{ -x^2 }{ 2 }} }{ 1-c e ^{\frac{ -x^2 }{ 2 }} }\]

Answer 10

I got y=1/(e^(x^2/2)-1) as the final answer.

Answer 11

Never mind. I got it!