solve the following diffrential equations(1+3v^2)/V(1-V^2)

Question

differential equations

Answer 1

Do you mean integrate ∫(1+3v²)dv/(v(1-v²))? The expression you have provided is not a differential equation. An equation would contain an equals sign and a differential equation would also contain at least one differential component such as dv/dx. You can always resubmit your question or amend it by editing.

I'll assume that's an integration.

(1+3v²)dv/(v(1-v²)) can be expressed in partial fractions which will usually simplify integration.

(1+3v²)/(v(1-v²))≡A/v+B/(1-v)+C/(1+v) because 1-v²=(1-v)(1+v). A, B, C are constants.

1+3v²≡A-Av²+Bv+Bv²+Cv-Cv². By matching coefficients:

(1) constant: A=1;

(2) v²: -A+B-C=3;

(3) v: B+C=0.

Substitute A=1 in (2): -1+B-C=3, so B-C=4;

Add this to (3): 2B=4, B=2, therefore C=-2.

(1+3v²)/(v(1-v²))≡A/v+B/(1-v)+C/(1+v)=1/v+2/(1-v)-2/(1+v).

Now the integral becomes:

∫dv/v+2∫dv/(1-v)-2∫dv/(1+v)=ln(v)-2ln(1-v)-2ln(1+v)+C, where C is the integration constant.

Note that 2ln(x) is the same as ln(x²) where x is any quantity, and -ln(x) is the same as ln(1/x).

Therefore, -2[ln(1-v)+ln(1+v)]=-2ln[(1-v)(1+v)]=-2ln(1-v²)=-ln(1-v²)²=ln(1/(1-v²)²).

ln(v/(1-v²)²)+C is the solution.