# ode differiential equation

• Aug 12th 2010, 06:01 AM
Monster32432421
ode differiential equation
Could someone solve these:
xyy'+y^2=sinx by letting y^2=v
y'=2/(x+2y-3) by letting x+2y-3=v
2y'y''=1+(y')^2 by letting y'=v
or explain one and if they are similar then i'd be fine?
• Aug 12th 2010, 07:17 AM
DeMath
Quote:

Originally Posted by Monster32432421
Could someone solve these:

xyy'+y^2=sinx

or explain one and if they are similar then i'd be fine?

Hint:

Multiply both sides of the equation by $\displaystyle 2x$

$\displaystyle 2x^2yy'+2xy^2=2x\sin{x}$

$\displaystyle (x^2y^2)'=2x\sin{x}$

$\displaystyle x^2y^2=2\int{x\sin{x}\,dx}$
• Aug 12th 2010, 07:31 AM
DeMath
Quote:

Originally Posted by Monster32432421
Could someone solve these:

2y'y''=1+(y')^2 by letting y'=v

or explain one and if they are similar then i'd be fine?

Hint:

$\displaystyle 2y'y''=1+(y')^2$

$\displaystyle \dfrac{2y'y''}{1+(y')^2}=1$

$\displaystyle \Bigl(\ln(1+(y')^2)\Bigl)'=1$

$\displaystyle \ln(1+(y')^2)=x+C$

$\displaystyle 1+(y')^2=e^{x+C}=C_1e^x$

$\displaystyle (y')^2=C_1e^x-1$

$\displaystyle y'=\pm\sqrt{C_1e^x-1}$

$\displaystyle y=\pm\int\!\sqrt{C_1e^x-1}\,dx$
• Aug 12th 2010, 07:37 AM
Monster32432421
Quote:

Originally Posted by DeMath
Hint:

Multiply both sides of the equation by $\displaystyle 2x$

$\displaystyle 2x^2yy'+2xy^2=2x\sin{x}$

$\displaystyle (x^2y^2)'=2x\sin{x}$

$\displaystyle x^2y^2=2\int{x\sin{x}\,dx}$

I actually forgot to write the substitute for it... sorry... it's y^2=v
• Aug 12th 2010, 07:54 AM
Monster32432421
Quote:

Originally Posted by DeMath
Hint:

$\displaystyle 2y'y''=1+(y')^2$

$\displaystyle \dfrac{2y'y''}{1+(y')^2}=1$

$\displaystyle \Bigl(\ln(1+(y')^2)\Bigl)'=1$

$\displaystyle \ln(1+(y')^2)=x+C$

$\displaystyle 1+(y')^2=e^{x+C}=C_1e^x$

$\displaystyle (y')^2=C_1e^x-1$

$\displaystyle y'=\pm\sqrt{C_1e^x-1}$

$\displaystyle y=\pm\int\!\sqrt{C_1e^x-1}\,dx$

how did you get from... $\displaystyle \dfrac{2y'y''}{1+(y')^2}=1$ to $\displaystyle \Bigl(\ln(1+(y')^2)\Bigl)'=1$
• Aug 12th 2010, 08:31 AM
Jester
If you let $\displaystyle y' = v$ then $\displaystyle y'' = v'$ and you get

$\displaystyle 2v \dfrac{dv}{dx} = 1 + v^2$ (separable).