Question

Solve the given initial-value problem

(3y + 2t - 5)dt + (4y+ 3t - 1)dy = 0, y(-1)=2

Answer 1

D.E is

$$(3y + 2t - 5)dt + (4y+ 3t - 1)dy = 0$$

Let,

$$y=y'+k\&t=t'+h$$

Such that

$$2h+3k-5=0\\3h+4k-1=0$$

Thus, $(h,k)=(13,-7)$ and actual equation reduced to homogeneous equation such that

$$\frac{dy'}{dt'}=-\dfrac{2t'+3y'}{3t'+4y'}$$

Let,

$$y'=vt'\implies v+t'\frac{dv}{dt'}=-\dfrac{2+3v}{3+4v}\\ \implies \frac{dt'}{t'}=-\dfrac{4v+3}{4v^2+6v+2}dv\\ \implies \ln|t'|=-\frac{1}{2}\ln|(2v^2+3v+1)|\\ \implies c t'^2=\frac{1}{2v^2+3v+1}\\ \implies ct'^4=\frac{1}{2y'^2+3y't'+t'^2}$$

But,

$$y'=y+7\&t'=t-13$$

Thus,

$$c(t-13)^4=\frac{1}{2(y+7)^2+3(y+7)(t-13)+(t-13)^2}$$

As, $y(-1)=2$ ,thus

$$c=\frac{1}{14^4(-20)}$$

Then the solution will be

$$-\frac{1}{20\cdot14^4}=\frac{1}{2(y+7)^2+3(y+7)(t-13)+(t-13)^2}$$