Differential EQs

[InterserveVB]

Can anyone help me with the following problems?

What is the solution to the differential eq dy/dx = (x^2)y^(1/2) with initial condition y(0) = 9?

Find a differential eq for dy/dx = (1+x^2)(1+y^2)

Also, if I am given dy/dt = t^2 / y how would I describe the resulting segments on a graph in terms of slope.

 

[Matt]

What is the solution to the differential eq dy/dx = (x^2)y^(1/2) with initial condition y(0) = 9?

Separation of variables.

Find a differential eq for dy/dx = (1+x^2)(1+y^2)

That is a differential equation.

Also, if I am given dy/dt = t^2 / y how would I describe the resulting segments on a graph in terms of slope.

Draw a slope field. Along the line y=0, the slope is undefined, so your diagram will consist of two pieces: the area above y=0 and the area below y=0.

 

[InterserveVB]

If I do sep of varibles I get dy / y^1/2 = x^3 / 3 where do I go from there?

 

[Matt]

If I do sep of varibles I get dy / y^1/2 = x^3 / 3 where do I go from there?

\(\displaystyle \L \begin{eqnarray}
\frac{dy}{y^{1/2}}&=&x^2\,dx\\
\int\frac{dy}{y^{1/2}}&=&\int x^2\,dx\\
2y^{1/2}&=&\frac{x^3}{3}+C
\end{eqnarray}\)

and solve for y.

 

[InterserveVB]

ok cool y = (x^3 / 6 +3)^2 thx

Also, for

Find a differential eq for dy/dx = (1+x^2)(1+y^2)

I forgot to put in terms of y =

I am down to dy / (1+ y^2) = x + (x^3/3) where do I go from here?

 

[Matt]

Also, for

Find a differential eq for dy/dx = (1+x^2)(1+y^2)

I forgot to put in terms of y =

I am down to dy / (1+ y^2) = x + (x^3/3) where do I go from here?

A handy integral to know: \(\displaystyle \L \int\frac{dy}{1+y^2}=\tan^{-1}y\) (plus a constant, of course).

 

[InterserveVB]

ah cool thx