Question

Find the general solution of each differential equation, expressing y explicitly in terms of x.

3) dy/dx= 3x²(y+1)

4) dy/dx = y-1 / 2x-1

Answer 1

(i)

$\frac{dy}{dx} = 3x²(y+1)$

=> $\frac{dy}{y+1} = 3x²dx$

Integrating we get

$\int\frac{dy}{y+1} = \int3x²dx$

=> ln |y+1| = x³ + ln |A| where ln |A| is integration constant

=> ln |y+1| - ln |A| = x³

=> $ln \mid {\frac {y+1}{A}}\mid = x³$

=> y + 1 = Ae^x³

=> y = Ae^x³ - 1

This is the solution of the given differential equation

(ii)

$\frac{dy}{dx} = \frac{y-1}{2x-1}$

=> $\frac{dy}{y-1} = \frac{dx}{2x-1}$

Integrating we get

=> $\int\frac{dy}{y-1} =\frac{1}{2}\int \frac{2dx}{2x-1}$

=> ln |y-1| = $\frac{1}{2} ln |2x-1| + ln |A|$

where ln |A| is integration constant

=> ln |y-1| - ln |A| = $\frac{1}{2} ln|2x-1|$

=> $ln \mid\frac{y-1}{A}\mid= ln{\sqrt{2x-1}}$

=> $\frac{y-1}{A}= \sqrt{2x-1}$

=> y - 1 = A$\sqrt{2x-1}$

=> y = 1 + A$\sqrt{2x-1}$

This is the solution of the given differential equation