Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • (i) The degree of the differential equation is d^2y/dx^2+e^dy/dx = 0 is ___

(i) The degree of the differential equation \frac{d^{2} y}{d x^{2}}+e^{\frac{d y}{d x}}=0 is
(ii) The degree of the differential equation \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=x is

(iii) The number of arbitrary constants in the general solution of a differential equation of order three is


(iv) \frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x} is an equation of the type

(v) General solution of the differential equation of the type \frac{d y}{d x}+P y=Q is given by

(vi) The solution of the differential equation  x \frac{d y}{d x}+2 y=x^{2} is

(vii) The solution of \left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0 is

(viii) The solution of the differential equation y d x+(x+x y) d y=0 is

(ix) General solution of \frac{d y}{d x}+y=\sin x is

(x) The solution of differential equation \cot y d x=x d y is

(xi) The integrating factor of \frac{d y}{d x}+y=\frac{1+y}{x}is

Answers (1)

(i) Given differential equation is
$$ \frac{d^{2} y}{d x^{2}}+e^{\frac{d y}{d x}}=0
Degree of this equation is not defined as it cannot be expresses as polynomial of derivatives.
(ii) We have \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=x
\Rightarrow 1+\left(\frac{d y}{d x}\right)^{2}=x^{2}
So, degree of this equation is two.

(iii) Given that the general solution of a differential equation has three arbitrary constants. So we require three more equations to eliminate these three constants. We can get three more equations by differentiating given equation three times. So, the order of the differential equation is three.

(iv) We have \frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}
The equation is of the type \frac{d y}{d x}+P y=Q
Hence it is linear differential equation.

(v) We have \frac{d x}{d y}+P_{1} x=Q_{1}
For solving such equation we multiply both sides by
So we get e^{\int P_{1} d y}\left(\frac{d x}{d y}+P_{1} x\right)=Q_{1} e^{\int P_{1} d y}
\\\Rightarrow \quad \frac{d x}{d y} e^{\int P_{1} d y}+P_{1} e^{\int P_{1}dy}=Q_{1} e^{\int P_{1}d y}$ \\\\$\Rightarrow \quad \frac{d}{d y}\left(x e^{\int P_{1} d y}\right)=Q_{1} e^{\int P_{1} dy}$ \\$\Rightarrow \quad \int \frac{d}{d y}\left(x e^{\int P_{1} d y}\right) d y=\int Q_{1} e^{P_{1} d y}dy$ \\$\Rightarrow \quad x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C
This is the required solution of the given differential equation.
(vi) We have, x \frac{d y}{d x}+2 y=x^{2}
\frac{d y}{d x}+\frac{2 y}{x}=x
This equation of the form \frac{d y}{d x}+P y=Q
\therefore \quad$ I.F. $=e^{\int \frac{2}{x} d x}=e^{2 \log x}=x^{2}
The general solution is
$$ y x^{2}=\int x \cdot x^{2} d x+C
\\\Rightarrow \quad y x^{2}=\frac{x^{4}}{4}+C$ \\$\Rightarrow \quad y=\frac{x^{2}}{4}+C x^{-2}

(vii) We have \left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0
\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{4 x^{2}}{1+x^{2}}
This equation is of the form \frac{d y}{d x}+P y=Q

$$ \therefore \quad \quad \quad \quad \mathrm{F} \cdot=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}
So, the general solution is:
$$ \begin{array}{ll} & y \cdot\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \frac{4 x^{2}}{\left(1+x^{2}\right)} d x+C \\ \Rightarrow & \left(1+x^{2}\right) y=\int 4 x^{2} d x+C \\ \Rightarrow & \left(1+x^{2}\right) y=\frac{4 x^{3}}{3}+C \\ \Rightarrow & y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}+C\left(1+x^{2}\right)^{-1} \end{array}
(viii) We have, y d x+(x+x y) d y=0$ $\Rightarrow \quad y d x+x(1+y) d y=0$
\\ \Rightarrow$ $\frac{d x}{-x}=\left(\frac{1+y}{y}\right) d y$ \\$\int \frac{1}{x} d x=-\int\left(\frac{1}{y}+1\right) d y$ \\$\Rightarrow \quad \log x=-\log y-y+\log C$ \\$\Rightarrow \quad \log x+\log y-\log C=-y$ \\$\Rightarrow \quad \log \frac{x y}{C}=-y$ \\$\Rightarrow \quad \frac{x y}{C}=e^{-y}
\Rightarrow \quad x y=C e^{-y}


(ix) We have, \frac{d y}{d x}+y=\sin x
Which is of the form \frac{d y}{d x}+P y=Q
$$ \text { I.F. }=e^{\int 1 d x}=e^{x}
So, the general solution is:
$$ \begin{array}{ll} & y \cdot e^{x}=\int e^{x} \sin x d x+C \\ \Rightarrow & y e^{x}=\frac{1}{2} e^{x}(\sin x-\cos x)+C \\ \Rightarrow & y=\frac{1}{2}(\sin x-\cos x)+C e^{-x} \end{array}

(x) Given differential equation is \cot y d x=x d y
\Rightarrow $$ \begin{array}{l} \int \frac{1}{x} d x=\int \tan y d y \\ \log x=\log \sec y+\log C \end{array}
\\\Rightarrow$ $ \quad \log \frac{x}{\sec y}=\log C$ \\$\frac{x}{\sec y}=C$ \\$\Rightarrow$ $x=C \sec y


(xi) Given differential equation is
$$ \begin{array}{l} \frac{d y}{d x}+y=\frac{1+y}{x} \\ \frac{d y}{d x}+y=\frac{1}{x}+\frac{y}{x} \end{array}
\Rightarrow \quad \frac{d y}{d x}+y\left(1-\frac{1}{x}\right)=\frac{1}{x}
Which is linear differential equation.
$$ \therefore \quad \text { I.F. }=e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x-\log x}=e^{x} \cdot e^{-\log x}=\frac{e^{x}}{x}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question