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UPSC 2015 Maths Optional Paper 1 Q5b — Step-by-Step Solution

10 marks · Section B

Exact equations · ODEs · asked 9× in 13 yrs · Read the full method →

Question

Solve the differential equation:

Technique

Exactness test; integrating factor depending on $y$ found via $(N_x-M_y)/M$ being a function of $y$ alone (the planted structure $M=y\cdot(\cdots)$ and $N_x-M_y=-4\cdot(\cdots)$ makes the ratio $-4/y$ clean); reconstruct potential $F$ by partial integration.

Solution

Strategy. Check exactness; if not exact, look for an integrating factor depending only on $y$ (or only on $x$).

Let $M=2xy^4 e^y+2xy^3+y$, $N=x^2 y^4 e^y-x^2 y^2-3x$.

Step 1 — Test exactness

$M_y\ne N_x$, not exact.

Step 2 — Integrating factor depending on $y$ alone

For $\mu=\mu(y)$:

Compute $N_x-M_y$:

Simplify $2xy^4 e^y-2xy^3 e^y(4+y)=2xy^3 e^y(y-(4+y))=2xy^3 e^y(-4)=-8xy^3 e^y$.

And $-2xy^2-6xy^2=-8xy^2$.

And $-3-1=-4$.

So $N_x-M_y=-8xy^3 e^y-8xy^2-4=-4(2xy^3 e^y+2xy^2+1)$.

Compare with $M=2xy^4 e^y+2xy^3+y=y(2xy^3 e^y+2xy^2+1)$.

So $\dfrac{N_x-M_y}{M}=\dfrac{-4(2xy^3 e^y+2xy^2+1)}{y(2xy^3 e^y+2xy^2+1)}=\dfrac{-4}{y}$.

Yes — function of $y$ alone. So IF is

Step 3 — Multiply by $\mu$

New $\tilde M=\dfrac{M}{y^4}=\dfrac{2xy^4 e^y+2xy^3+y}{y^4}=2xe^y+\dfrac{2x}{y}+\dfrac{1}{y^3}$.

New $\tilde N=\dfrac{N}{y^4}=\dfrac{x^2 y^4 e^y-x^2 y^2-3x}{y^4}=x^2 e^y-\dfrac{x^2}{y^2}-\dfrac{3x}{y^4}$.

Step 4 — Verify exactness of $(\tilde M,\tilde N)$

Equal ✓. Exact.

Step 5 — Find potential $F$

$F_x=\tilde M=2xe^y+\dfrac{2x}{y}+\dfrac{1}{y^3}$.

Integrate w.r.t. $x$:

Differentiate w.r.t. $y$ and compare with $\tilde N$:

Compare with $\tilde N=x^2 e^y-\dfrac{x^2}{y^2}-\dfrac{3x}{y^4}$: $g'(y)=0$, so $g(y)=$ const.

Step 6 — General solution

Answer