Question

Answer the following by appropriately matching the lists based on the information given in Column $\text{I}$ and Column $\text{II}$
$\begin{array}{cc}ColumnI& ColumnII\\ \text{a.}\text{If the function}y=e^{4x}+2e^{-x}\text{is a solution of}& \\ \text{the differential equation}\frac{\frac{d^{3}y}{d{x}^3}-13\frac{dy}{dx}}{y}=K,& \\ \text{then the value of}\frac{K}{3}\text{is}& \text{p.}3\\ \text{b.}\text{Number of straight lines which satisfy the}& \\ \text{differential equation}\frac{dy}{dx}+x{\left(\frac{dy}{dx}\right)}^2-y=0& \\ \text{is}& \text{q.}4\\ \text{c.}\text{If real value of}m\text{for which the substitution,}& \\ y=u^m\text{will transform the differential equat-}& \\ \text{ion,}2x^{4}y\frac{dy}{dx}+y^4=4x^6\text{into a homogeneous}& \\ \text{equation, then the value of}2m\text{is}& \text{r.}2\\ \text{d.}\text{If the solution of differential equation}& \\ x^2\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}=12y\text{is}y=A{x}^m+B{x}^{-n},& \\ \text{then}|m+n|\text{is}& \text{s.}7\end{array}$

Then which of the following is correct ?

• A. $a\to s,b\to r,c\to p,d\to q$
• B. $a\to q,b\to r,c\to p,d\to s$
• C. $a\to q,b\to p,c\to r,d\to s$
• D. $a\to s,b\to q,c\to p,d\to s$

Answer 1

The correct option is B $a\to q,b\to r,c\to p,d\to s$

a. $y=e^{4x}+2e^{-x};y_1=4e^{4x}-2e^{-x};y_2=16e^{4x}+2e^{-x};y_3=64e^{4x}-2e^{-x}$
Now, $y_3-13y_1=(64e^{4x}-2e^{-x})-13(4e^{4x}-2e^{-x})$
$=12e^{4x}+24e^{-x}$
$y_3-13y=12(e^{4x}+2e^{-x})=12y$
$\therefore K=12$ and $\frac{K}{3}=4$

b. Since equation is of degree 2, two lines are possible.

c. $y=u^m$ or $\frac{dy}{dx}=m{u}^{m-1}\frac{du}{dx}$
Substituting the value of $y$ and $\frac{dy}{dx}$ in $2x^{4}y\frac{dy}{dx}+y^4=4x^6$,
we have
$2x^4{u}^{m}m{u}^{m-1}\frac{du}{dx}+u^{4m}=4x^6$ or $\frac{du}{dx}=\frac{4x^6-u^{4m}}{2m{x}^4{u}^{2m-1}}$
For homogeneous equation, $4m=6$ or $m=\frac{3}{2}$
and $2m-1=2$ or $m=\frac{3}{2}$.

d.
$y=A{x}^m+B{x}^{-n}$
$\therefore \frac{dy}{dx}=Am{x}^{m-1}-nB{x}^{-n-1}$
$\therefore \frac{d^{2}y}{d{x}^2}=Am(m-1)x^{m-2}+n(n+1)B{x}^{-n-2}$
Putting these values in $x^2\frac{d^{2}y}{d{x}^2}+2x\frac{dy}{dx}=12y$, we get
$m(m+1)A{x}^m+n(n-1)B{x}^{-n}=12(A{x}^m+B{x}^{-n})$
i.e., $m(m+1)=12$ or $n(n-1)=12$
i.e., $m=3,-4$ or $n=4,-3$