A) circles
B) parabolas
C) ellipses
D) hyperbolas
Correct Answer: D
Solution :
The given differential equation is (3x + 4y + 1)dx + (4x + 5y + 1) dy = 0 ....(i) Comparing Eq. (i) with Mdx + Ndy = 0, we get M = 3x + 4y + 1 and N = 4x + 5y + 1 Here, \[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=4\] Hence, Eq. (i) is exact and solution is given by \[\int{\left( 3x+4y+1 \right)dx+\int{\left( 5y+1 \right)dy=C}}\] \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\] \[\frac{3{{x}^{2}}}{2}+4xy+x+\frac{5{{y}^{2}}}{2}+y-C=0\] \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\] \[3{{x}^{2}}+8xy+2x+5{{y}^{2}}+2y-2C=0\] \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\] \[3{{x}^{2}}+2.4xy+2x+5{{y}^{2}}+2y+C'=0\] ..(ii) where, \[C'=-2C\] On comparing Eq. (ii) with standard form of conic section \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+C=0\] We get, a = 3 h = 4 b = 5 Here, \[{{h}^{2}}-ab=16-15=1>0\] Hence, the solution of differential equation represents family of hyperbolas.
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