Mathematics 2 Question. Please help! Description below! PLEASE!?

You did a fine job. You just need to proceed, calculating both integrals. The first integral can be done through dividing up in partial fractions : (1-2v)/(v(1+5v)) = A/v + B/(1+5v) (1-2v) = A(1+5v) + Bv A = 1 and 5A+B = -2 => B = -7 So we have integral (1/v - 7/(1+5v)) dv ln(|v|) - (7/5) ln(|1+5v|) + C The second integral is just ln(|x|) + C' .

So we have ln(|x|) + C" = ln(|v|) - (7/5) ln(|1+5v|) ln(|v|) + ln(|1+5v|^(-7/5)) ln(|v (1+5v)^(-7/5)|) = ln(C# |x|) (C# > 0) Now take exp(...) of both sides of the last equation C# *|x| = | v(1+5v)^(-7/5) Now take both sides to the 5th power C#^5 |x^5| = | v(1+5v)^-7 Now raise to the -1/7th power C#^(-5/7) |x^-5/7| = |v(1+5v) Name C#^(-5/7) = K > 0 K |x^-5/7| = |v(1+5v) We do a sign investigation of v(1+5v) : ....................-1/5..........0....... v : -------------------------0.

SEE MY TRIAL. .... I DON'T KNOW WHAT TO DO NEXT .... OR AM WRONG SOMEWHERE. ... Solve dy/dx=2xy+3y^2/x^2-2xy.

SOLUTION dy/dx = (2xy+3y²)/(x²-2xy) Divide both numerator and denominator by x² dy/dx = (2xy+3y²)/x²/(x²-2xy)/x² dy/dx = 2(y/x) + 3(y/x)²/1 - 2(y/x) Using: y = vx , v = y/x v = y/x dy/dx = v + x(dv/dx) Therefore, dy/dx = 2(y/x) + 3(y/x)²/1 - 2(y/x) v + x(dv/dx) = (2v + 3v²)/(1 - 2v) x(dv/dx) = (2v + 3v²)/(1 - 2v) - v x(dv/dx) = (2v + 3v²) - v(1 - 2v)/(1 - 2v) x(dv/dx) = 2v + 3v² - v + 2v²/(1 - 2v) x(dv/dx) = (v + 5v²)/(1 - 2v) Using separable variable .. cross multiply to get.. xdv(1 - 2v) = (v + 5v²)dx This can be re write as ... (1 - 2v)/(v + 5v²)dv = dx/x Put the sign of integration on both side.?(1 - 2v)/(v + 5v²)dv =? Dx/x?(1 - 2v)/(v + 5v²)dv =?(1/x)dx I DON'T KNOW WHAT TO DO AGAIN. ..

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