UPSC Maths 2013 Paper 1 Q6a — Solution

10 marks · Section B

Question

Solve the differential equation

(5x^{3}+12x^{2}+6y^{2})\,dx+6xy\,dy=0.

Standard test-and-fix integrating-factor method for first-order ODEs in differential form: $(M_y-N_x)/N$ pure- $x$ ⇒ IF is $\exp\int(\cdot)dx$ .

Strategy. Check exactness; if not exact, find an integrating factor; then solve as exact.

With $M=5x^{3}+12x^{2}+6y^{2}$ and $N=6xy$ :

\frac{\partial M}{\partial y}=12y,\qquad\frac{\partial N}{\partial x}=6y.

Not exact ( $M_y\ne N_x$ ).

Test whether $(M_y-N_x)/N$ depends on $x$ only:

\frac{M_y-N_x}{N}=\frac{12y-6y}{6xy}=\frac{6y}{6xy}=\frac{1}{x}.

Depends only on $x$ ✓. Integrating factor:

\mu(x)=\exp\!\int\frac{1}{x}\,dx=x.

$(5x^{4}+12x^{3}+6xy^{2})\,dx+6x^{2}y\,dy=0$ .

New $M'=5x^{4}+12x^{3}+6xy^{2}$ , $N'=6x^{2}y$ . Check:

\frac{\partial M'}{\partial y}=12xy=\frac{\partial N'}{\partial x}\;\checkmark.

Exact.

Integrate $F_y=6x^{2}y$ with respect to $y$ :

F(x,y)=3x^{2}y^{2}+g(x).

Differentiate w.r.t. $x$ and match $F_x=M'$ :

F_x=6xy^{2}+g'(x)\stackrel{!}{=}5x^{4}+12x^{3}+6xy^{2}\;\Longrightarrow\;g'(x)=5x^{4}+12x^{3}.

Integrate: $g(x)=x^{5}+3x^{4}$ .

Therefore $F(x,y)=x^{5}+3x^{4}+3x^{2}y^{2}$ .

\boxed{\;x^{5}+3x^{4}+3x^{2}y^{2}=C.\;}