Question

y'=(4x+3y+15)/(2x+y+7)

Answer 1

$\frac{dy}{4x+3y+15}=\frac{−dx}{2x+y+7}=\frac{dy−λdx}{4x+3y+15+λ(2x+y+7)}$

$=\frac{d(y−λx)}{(λ+3)y+(2λ+4)x+7λ+15}$

$=\frac{d((λ+3)y−λ(λ+3)x)}{(λ+3){(λ+3)y+(2λ+4)x+7λ+15}}$

For consistency in numerator and denominator

$−λ(λ+3)=2λ+4⟹$

$λ2+5λ+4=0⟹λ=−1or−4$

Equating two ratios with the two $λ ’s$

$\frac{d((−1+3)y−(−1)(−1+3)x)}{(−1+3){(−1+3)y+(2⋅−1+4)x+7⋅−1+15}}=\frac{d((−4+3)y−(−4)(−4+3)x)}{(−4+3){(−4+3)y+(2⋅−4+4)x+7⋅−4+15}}$

$\frac{d(2y+2x+8)}{2{2y+2x+8}}=\frac{d(−y−4x−13)}{−1{−y−4x−13}}$

$\frac{d(y+x+4)}{{y+x+4}}+2\frac{d(y+4x+13)}{{y+4x+13}}=0$

Integrating,

$ln(y+x+4)+2ln(y+4x+13)+lnC$

$(y+x+4)(y+4x+13)2=C$

Verification

since my answer is at variance to other answers it needs verification.

Differentiating (1)

$2(y+x+4)(dy+4dx)+(y+4x+13)(dy+dx)=0$

$(2y+2x+8+y+4x+30dy+(8y+8x+32++y+4x+13)dx=0$

$(6x+3y+21)dy+(12x+9y+45)dx=0$

$(2x+y+7)dy+(4x+3y+15)dx=0$

Verified.