Solving ODE: How to Integrate Left Integral?

[hanson]

Hi all!
How to solve this ODE?
$$\frac{a+b-c\sqrt{H}}{k}=\frac{dH}{dt}$$ , where a,b, c and k are constants; H is the variable

I am up to this step:
$$\int \frac{dH}{a+b-c\sqrt{H}}=\int \frac{dt}{k}$$
and I don't know how to integrate the left integral, can anyone help please?

 

[Ali 2]

Hi,


$$I = \int \frac { dH} { a + b- c \sqrt H }$$

Let u²= H >>> 2u du = dH

$$\therefore I = \int \frac {2u } { a + b- c u } du = \frac {-2} {c} \int \frac { -cu + a + b- (a + b) } { a + b- cu } du = = \frac {-2} {c} \left( 1 - \frac { a + b} { -c } \int \frac { -c } { a + b- cu } du \right)$$

$$= \frac { -2 } {c} u - \frac { 2(a + b) } {c^2} \ln | a + b- c u | + C = \frac { -2 } {c} \sqrt {H} - \frac { 2(a + b) } {c^2} \ln | a + b- c \sqrt {H} | + C$$

 

[hanson]

Hi,


$$I = \int \frac { dH} { a + b- c \sqrt H }$$

Let u²= H >>> 2u du = dH

$$\therefore I = \int \frac {2u } { a + b- c u } du = \frac {-2} {c} \int \frac { -cu + a + b- (a + b) } { a + b- cu } du = = \frac {-2} {c} \left( 1 - \frac { a + b} { -c } \int \frac { -c } { a + b- cu } du \right)$$

$$= \frac { -2 } {c} u - \frac { 2(a + b) } {c^2} \ln | a + b- c u | + C = \frac { -2 } {c} \sqrt {H} - \frac { 2(a + b) } {c^2} \ln | a + b- c \sqrt {H} | + C$$

Many thanks.
But it is very hard to have a closed form of H in terms of t for the solution then?

 

[Ali 2]

It is not necessary to have an explicit relation, implicit relation is sufficuint

 

[hanson]

oops..sorry..
I have made a mistake in modelling...
The differential equation should be:
$$\frac{a+bsinwt-c\sqrt{H}}{k}=\frac{dH}{dt}$$
which again I don't know how to solve...
I simply can't separate it...
please help..

 

[blumfeld0]

I don't know how to do this one but I know the answer using Mathematica

H[t]= a t w/k - c t w/k H^1/2 + k w/k (Constant of integration)- b Cos[wt]/k

blumfeld0

 

[hanson]

Hi blumfeld0,
is your answer for the amended ODE I just posted?

 

[hanson]

to confirm, is the solution:
$$H(t)=\frac{atw}{k}-ct\frac{w}{k}\sqrt{H}+\frac{kw}{k}(constant of integration)-bcos\frac{wt}{k}$$

but the "k" can be cancelled?

 

[hanson]

can anyone help confirm the answer?

 

[saltydog]

Mathematica 5.2 cannot solve this:

[tex]
\text{DSolve[}h^{'}[t]==\frac{a+b Sin[\omega t]-c\sqrt{h[t]}}{k},h,t]
[/tex]

So Blumfeld, how did you arrive at that expression?

Also, when I back-substitute yours into the ODE, Mathematica does not indicate it satisfies the ODE.