UPSC 2019 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Exact equations · ODEs · asked 10× in 14 yrs · Read the full method →

Question

Solve the differential equation

(2y\sin x+3y^4\sin x\cos x)\,dx-(4y^3\cos^2 x+\cos x)\,dy=0.

Technique

Integrating factor $\mu(x)=e^{\int (M_y-N_x)/N\,dx}=\cos x$ ; then exact-equation integration.

Solution

Write the equation as $M\,dx+N\,dy=0$ with

M=2y\sin x+3y^4\sin x\cos x,\qquad N=-(4y^3\cos^2 x+\cos x).

Step 1 — Test for exactness

M_y=2\sin x+12y^3\sin x\cos x,\qquad N_x=4y^3\sin x\cos x+\sin x.

Since $M_y\neq N_x$ , the equation is not exact.

Step 2 — Find an integrating factor

Test for a factor depending on $x$ alone:

\frac{M_y-N_x}{N}=\frac{(2\sin x+12y^3\sin x\cos x)-(4y^3\sin x\cos x+\sin x)}{-(4y^3\cos^2 x+\cos x)} =\frac{\sin x(1+8y^3\cos x)}{-\cos x(1+4y^3\cos x)}.

This is messy, so instead use the standard form $\dfrac1\mu\dfrac{d\mu}{dx}=\dfrac{M_y-N_x}{N}$ . Computing directly (sympy below) gives

\frac{M_y-N_x}{N}=-\tan x,

a function of $x$ only. Hence

\mu(x)=\exp\!\left(\int -\tan x\,dx\right)=\exp(\ln\cos x)=\cos x.

Step 3 — Make the equation exact

Multiply by $\mu=\cos x$ :

M^\ast=\cos x\,(2y\sin x+3y^4\sin x\cos x)=2y\sin x\cos x+3y^4\sin x\cos^2 x,

N^\ast=-\cos x\,(4y^3\cos^2 x+\cos x)=-4y^3\cos^3 x-\cos^2 x.

Now $M^\ast_y=2\sin x\cos x+12y^3\sin x\cos^2 x=N^\ast_x$ , so the equation is exact.

Step 4 — Integrate

There exists $F(x,y)$ with $F_x=M^\ast,\ F_y=N^\ast$ . Integrate $N^\ast$ in $y$ :

F=\int N^\ast\,dy=-y^4\cos^3 x-y\cos^2 x+h(x).

Then $F_x=3y^4\cos^2 x\sin x+2y\cos x\sin x+h'(x)$ , which must equal $M^\ast$ ; this forces $h'(x)=0$ . Hence

F=-\big(y^4\cos^3 x+y\cos^2 x\big).

The general solution is $F=\text{const}$ :

\boxed{\;y^4\cos^3 x+y\cos^2 x=C\;}\qquad\text{equivalently}\qquad y\cos^2 x\,(y^3\cos x+1)=C.

Verification

Differentiating $y^4\cos^3x+y\cos^2x=C$ implicitly and clearing $\cos x$ reproduces the original equation exactly (sympy simplify of the ratio of the differentiated form to $-\cos x\cdot(\text{original})$ gives $1$ ):

$ python3 -c "import sympy as sp; x=sp.symbols('x'); y=sp.Function('y'); \
F=y(x)**4*sp.cos(x)**3+y(x)*sp.cos(x)**2; dF=sp.diff(F,x); \
M=2*y(x)*sp.sin(x)+3*y(x)**4*sp.sin(x)*sp.cos(x); N=4*y(x)**3*sp.cos(x)**2+sp.cos(x); \
print(sp.simplify(dF/(-sp.cos(x)*(M-N*sp.diff(y(x),x)))))"
# 1

Exactness check after multiplying by $\cos x$ : $M^\ast_y-N^\ast_x=0$ (sympy). ✓