# Math Help - ode differiential equation

1. ## 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?

2. 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 $2x$

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

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

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

3. 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:

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

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

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

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

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

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

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

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

4. Originally Posted by DeMath
Hint:

Multiply both sides of the equation by $2x$

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

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

$x^2y^2=2\int{x\sin{x}\,dx}$
I actually forgot to write the substitute for it... sorry... it's y^2=v

5. Originally Posted by DeMath
Hint:

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

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

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

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

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

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

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

$y=\pm\int\!\sqrt{C_1e^x-1}\,dx$
how did you get from... $\dfrac{2y'y''}{1+(y')^2}=1$ to $\Bigl(\ln(1+(y')^2)\Bigl)'=1$

6. If you let $y' = v$ then $y'' = v'$ and you get

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