UPSC 2018 Maths Optional Paper 1 Q8d — Step-by-Step Solution

12 marks · Section B

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

Question

Find $f(y)$ such that $(2xe^y+3y^2)\,dy+(3x^2+f(y))\,dx=0$ is exact and hence solve.

Impose $P_y=Q_x$ to solve $f'(y)=2e^y$ ; then integrate the exact equation ( $F=\int P\,dx$ , match $F_y=Q$ ).

Write the equation in the standard order $P\,dx+Q\,dy=0$ with

P=3x^2+f(y),\qquad Q=2xe^{y}+3y^2.

The equation is exact iff $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ :

\frac{\partial P}{\partial y}=f'(y),\qquad \frac{\partial Q}{\partial x}=2e^{y}.

So we need

f'(y)=2e^{y}\;\Rightarrow\;f(y)=2e^{y}

(taking the integration constant to be $0$ , since any constant added to $f$ just shifts $P$ by a constant and can be absorbed).

\boxed{\;f(y)=2e^{y}.\;}

P=3x^2+2e^{y},\qquad Q=2xe^{y}+3y^2,\qquad P_y=2e^{y}=Q_x.\ \checkmark

Integrate $P$ in $x$ :

F=\int(3x^2+2e^{y})\,dx=x^{3}+2xe^{y}+g(y).

Differentiate in $y$ and match $Q$ :

F_y=2xe^{y}+g'(y)=Q=2xe^{y}+3y^2\;\Rightarrow\;g'(y)=3y^2\;\Rightarrow\;g(y)=y^{3}.

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