Solve the intial value problem.?

Question

(t^2 - 11t + 30)*dy = dt ; (t > 6), x(7) = 0

Answer 1

dy = dt/(t^2-11t+30) = (1/(t-6) - 1/(t-5)) dt

So, y = ln{(t-6)/(t-5)} + c

y(7) = 0

Thus, c = -ln(1/2) = ln 2

or y = ln{(t-6)/(t-5)} + ln 2

Answer 2

pleased with the intention to make issues less complicated, enable a = (y^2 - 2x)/(x^2) so we be conscious then, that the equation would be written as (y')^2 + 2a(y') - a = 0 so this would possibly not component out good however the roots are (making use of the quadratic formulation): -a + sqrt(a^2 -a) = U and -a - sqrt(a^2 - a) = V this means that the differential answer will resolve y' = U and y' = V then use the preliminary situations from there you ought to use a thank you to resolve non-homogeneous ODE's nonetheless it might desire to be a sprint hard because of the fact it does not look separable...it form of feels we are nonetheless left with a ''y' being prolonged via x in spite of if we strive to component out a 'y'...i wish that helped slightly! i wish that helped