Mathematics Test - 8

Concept:

The order of differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative accruing in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given:

\({\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^{\frac{1}{2}}} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^{\frac{1}{3}}}\)

Highest derivative in the given differential equation is 4

Hence order is 4.

To find the degrees, we need to change the differential equation in to form which is free from radicals.

\({\left[ {{{\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)}^{\frac{1}{2}}}} \right]^6} = {\left[ {{{\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]}^{\frac{1}{3}}}} \right]^6}\)

\(\Rightarrow {\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^3} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^2}\)

Now the differential equation is free from radicals

Degree of highest derivative = 3

∴ Order and degree of the given differential equation = 4 and 3 respectively.

Concept:

Differential Equation: A differential equation is an equation that relates one or more functions and their derivatives.

e.g. \(\rm \dfrac{dy}{dx}\) + x = 2y + 3, etc.

\(\rm \dfrac{d}{dx}x^n\) = nx^n-1.

Calculation:

The general equation of all lines passing through the origin is y = mx, where m is a constant. Differentiating this equation with respect to x, we get:

\(\rm \dfrac{d}{dx}(y)=\dfrac{d}{dx}(mx)\)

⇒ \(\rm \dfrac{dy}{dx}=m=\dfrac{y}{x}\)

∴ The answer is none of these.

Concept:

Separation of variable:

If the equation is such that the variables can be separated, then

• Separate the variables.
• Take a single variable on either side.
• Integrate both sides w.r.t the respective variable.

Calculation:

The given differential equation is 

\(\rm {dy\over dx}= {\sqrt{1-y^2}\over \sqrt{1-x^2}}\)

\(\rm {dy\over \sqrt{1-y^2}}= {dx\over\sqrt{1-x^2}}\)

Integrating both sides, we get

\(\rm \int{dy\over \sqrt{1-y^2}}= \int{dx\over\sqrt{1-x^2}}\)

⇒ sin^-1 y = sin^-1x + c

⇒ sin-1 y = sin^-1 x + sin^-1 c

⇒ sin-1 y - sin-1 x = sin-1 c

Concept:

Integrating factor of \(\frac{{dy}}{{dx}} + Py = Q\left( x \right)\) is given by,

IF = e^{∫ Pdx}

Calculation:

P = 2 / sin x

IF = e^{∫ 2 / sin x dx} = e^{∫ 2 cosec xdx}

IF = e^{2 log (tan (x / 2))}

∴ IF = tan² (x / 2)

Concept:

In first order linear differential equation;

\(\rm {dy\over dx}+Py=Q\), where P and Q are function of x

Integrating factor (IF) = e∫ P dx

General solution: y × (IF) = ∫ Q(IF) dx

Calculation:

The given equation is

(2y + x) \(\rm dy\over dx\) = 1

⇒ (2y + x) = \(\rm dx\over dy\)

⇒ \(\rm dx\over dy\) - x = 2y

∴ It is linear differential equation is of first order

IF = e∫ -1 dy

⇒ IF = e-y

Now, x × (IF) = ∫ Q (IF) dy

⇒ x × e-y = ∫ 2y × e-y dy

⇒ xe^-y = 2\(\rm \left[y\int e^{-y}dy - \int\left\{ {dy\over dy}\times\int e^{-y}dy\right\}dy\right]\)

⇒ xe-y = 2\(\rm \left[-ye^{-y} + \int e^{-y}dy\right]\) + c

⇒ xe-y = 2\(\rm \left[-ye^{-y} - e^{-y}\right]\) + c

⇒ x + 2y + 2 = ce^y

Concept:

The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x.

The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(ydx = \left( {y - x} \right)dy\)

\(y\frac{{dx}}{{dy}} = y - x\)

\(\frac{{dx}}{{dy}} + \frac{x}{y} = 1\)

It is form of \(\frac{{dx}}{{dy}} + Px = Q\)

\(I.F. = {e^{\smallint pdy}}\)

\(I.F. = {e^{lny}} = y\)

The solution of the linear equation is given by

\(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dy + c.\)

\(x\left( y \right) = \smallint 1\left( y \right)dy + c\)

\(xy = \frac{{{y^2}}}{2} + c\)

Concept:

For solving first order, first-degree differential equations always first inspect with variable separation method.

Calculation:

Given the differential equation is,

\(\frac{{dy}}{{dx}} + 7{x^2}y = 0 \Rightarrow \frac{{dy}}{{dx}} = - 7{x^2}y\), separating variables

\(\frac{{dy}}{y} = - 7{x^2}dx\), integrating both sides;

\(\smallint \frac{{dy}}{y} = - 7\smallint {x^2}dx;lny = - 7\frac{{{x^3}}}{3} + lnA\)

Where A is a constant.

\(\ln y - \ln A = - \frac{{7{x^3}}}{3} \Rightarrow \ln \left( {\frac{y}{A}} \right) = - \frac{{7{x^3}}}{3}\)

\(\frac{y}{A} = {e^{ - \frac{7}{3}{x^3}}} \Rightarrow y = A{e^{ - \frac{7}{3}{x^3}}}\) …(1)

Use condition \(y\left( 0 \right) = \frac{3}{7}\) in (1)

\(\Rightarrow \frac{3}{7} = A\;use\;in\;(1) \Rightarrow y = \frac{3}{7}{e^{ - \frac{{7{x^3}}}{3}}}\)

Key Points

Practice all the methods of solving first order first-degree differential equations.

Concept:  

\(\rm \int e^{x}dx = e^{x}\)

ln e^y = y .

Calculation: 

We have , 

dy = ex + 2y dx 

⇒ dy = e^x . e^2y dx 

⇒ e^-2y dy = e^x dx 

On Integrating both sides, We get

\(\rm -\frac{1}{2} \) e^-2y  = e^x + C          .... (i) 

It is given that y (0) = 0 , i.e. at x = 0 , y = 0 , putting this in (i) , 

\(\rm -\frac{1}{2} \) = 1 + C 

⇒ C = \(\rm -\frac{3}{2} \) . 

Putting C = \(\rm -\frac{3}{2} \) in (i) , we get 

\(\rm -\frac{1}{2} \) e^-2y = e^x  \(\rm -\frac{3}{2} \)

⇒ e^-2y = - 2e^x  + 3

⇒ e^2y = \(\rm \frac{1}{3- 2\ e^{x}} \) 

⇒ 2y = ln \(\rm \left ( \frac{1}{3- 2\ e^{x}} \right )\) 

⇒ y = \(\rm \frac{1}{2}\ln\left ( \frac{1}{3- 2\ e^{x}} \right )\) . 

The correct option is 3. 

Concept:

A differential equation is solved by separating the variables and then eliminating the differential terms by integrating both sides.

Calculation:

Given, \(x\frac{dy}{dx}=y(\log y-\log x+1)\)

⇒ \(\frac{dy}{dx}={y\over x}(\log {y\over x}+1)\)     -----(i)

Let y = v.x ⇒ \({dy \over dx} = v + x {dv \over dx}\)

Putting the value of y and \({dy \over dx}\) in (i),

⇒ \(v + x {dv \over dx} = v (\log v + 1)\)

⇒ \(x {dv \over dx} = v \log v\)

⇒ \({dv \over v \log v} ={dx \over x}\)

Integrating both sides,

⇒ \(\int{dv \over v \log v} =\int{dx \over x} \)

If log v = t ⇒ dv/v = dt

⇒ \(\int{dt \over t} = \log x + c\)

⇒ \(\log t = \log x + c\)

⇒ \(\log {t \over x} = c\)

⇒ t/x = e^c

⇒ t/x = C

⇒ t = Cx

⇒ \(\log v = Cx\)

⇒ \(\log {y \over x}= Cx\)

∴ The correct answer is option (4).

Concept:

\(\rm \int \frac{1}{x^2}\;dx = \frac{-1}{x} + c\)

Calculation:

Given:

x²dy + y²dx = 0

⇒ x²dy = -y²dx

\(\rm ⇒ \frac{dy}{y^2} = -\rm\frac{dx}{x^2}\)

Integrating both sides, we get

\(\rm \int \frac{dy}{y^2} = -\rm\int\frac{dx}{x^2}\)

 \(\rm \frac{-1}{y} = -\frac{-1}{x} + c\\\rm \frac{-1}{y} = \frac{1}{x} + c\\ \frac{1}{x}+ \frac{1}{y}= -c\\x+y=-cxy\)

Here c is differential constant, take -c = 1/c (Because 1/c is also a constant)

⇒ c(x + y) = xy