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Differential Equation and Applications (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differential Equation and Applications (p-1)5 Marks
Question
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:
SolutionD.E.
(i)xy=log y+ky^(')(1-xy)=y^(2)
(ii)y=x^(n)x^(2)(d^(2)y)/(dx^(2))-nx(dy)/(dx)+ny=0
(iii)y=e^(x)(dy)/(dx)=y
(iv)y=1-log xx^(2)(d^(2)y)/(dx^(2))=1
(v)y=ae^(x)+be^(-x)(d^(2)y)/(dx^(2))=y
(vi)ax^(2)+by^(2)=5xy(d^(2)y)/(dx^(2))+x((dy)/(dx))^(2)=y*(dy)/(dx)
✓

Answer

(i) $x y=\log y+k$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& x \cdot \frac{d y}{d x}+y \times 1=\frac{1}{y} \cdot \frac{d y}{d x}+0 \\
& \therefore x \frac{d y}{d x}+y=\frac{1}{y} \cdot \frac{d y}{d x} \\
& \left(x-\frac{1}{y}\right) \frac{d y}{d x}=-y
\end{aligned}
$
$\therefore\left(\frac{x y-1}{y}\right) \frac{d y}{d x}=-y$
$\therefore \frac{d y}{d x}=\frac{-y^2}{x y-1}=\frac{y^2}{1-x y}$, if $x y \neq 1$
$\therefore y^{\prime}=\frac{y^2}{1-x y^{\prime}}$, if $x y \neq 1$.
$\therefore y^{\prime}(1-x y)=y^2$
Hence, $x y=\log y+k$ is a solution of the D.E. $y^{\prime}(1-x y)=y^2$.
(ii) $y=x^n$
Differentiating twice w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^n\right)=n x^{n-1} \\
& \text { and } \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(n x^{n-1}\right)=n \frac{d}{d x}\left(x^{n-1}\right)=n(n-1) x^{n-2} \\
& \begin{aligned}
\therefore x^2 \frac{d^2 y}{d x^2}-n x \frac{d y}{d x}+n y \\
& =x^2 \cdot n(n-1) x^{n-2}-n x \cdot n x^{n-1}+n \cdot x^n \\
& =n(n-1) x^n-n^2 x^n+n x^n \\
& =\left(n^2-n-n^2+n\right) x^n=0
\end{aligned}
\end{aligned}
$
This shows that $y=x^n$ is a solution of the D.E.
$
x^2 \frac{d^2 y}{d x^2}-n x \frac{d y}{d x}+n y=0
$
(iii) $y = e ^{ x }$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}= e ^{ x }= y
$
Hence, $y = ex$ is a solution of the D.E. $\frac{d y}{d x}= y$.
(iv) $y=1-\log x$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}(1)-\frac{d}{d x}(\log x) \\
& =0-\frac{1}{x}=-\frac{1}{x}
\end{aligned}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d^2 y}{d x^2}=-\frac{d}{d x}\left(x^{-1}\right)=-(-1) x^{-2}=\frac{1}{x^2} \\
& \therefore x^2 \frac{d^2 y}{d x^2}=1
\end{aligned}
$
Hence, $y=1-\log x$ is a solution of the D.E.
$
x^2 \frac{d^2 y}{d x^2}=1
$
(v) $y=a e^x+b e^{-x}$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}= a \left( e ^{ x }\right)+ b \left(- e ^{- x }\right)= ae ^{ x }- be ^{- x }
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d^2 y}{d x^2}= a \left( e ^{ x }\right)- b \left(- e ^{- x }\right) \\
& = ae ^{ x }+ be ^{- x }
\end{aligned}
$
$=y$
Hence, $y = ae ^{ x }+ be ^{- x }$ is a solution of the D.E. $\frac{d^2 y}{d x^2}= y$.
(vi) $a x^2+b y^2=5$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& a(2 x)+b\left(2 y \frac{d y}{d x}\right)=0 \\
& \therefore a x+b y \frac{d y}{d x}=0
\end{aligned}
$
$
\therefore a x=-b y \frac{d y}{d x}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& a \cdot 1=-b\left[y \frac{d}{d x}\left(\frac{d y}{d x}\right)+\frac{d y}{d x} \cdot \frac{d y}{d x}\right] \\
& \therefore a=-b\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]
\end{aligned}
$
Dividing (1) by (2), we get
$
\begin{aligned}
& x=\frac{y \frac{d y}{d x}}{y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2} \\
& \therefore x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2=y \frac{d y}{d x}
\end{aligned}
$
Hence, $a x^2+b y^2=5$ is a solution of the D.E.
$
x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2=y\left(\frac{d y}{d x}\right)
$

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