Solving Homogeneous Differential Equation MCQ Quiz - Objective Question with Answer for Solving Homogeneous Differential Equation - Download Free PDF

Calculation

Given equation: \(x\frac{dy}{dx} = y + x \tan(\frac{y}{x})\)

Divide by x: \(\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})\)

Let y = vx, then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

Substitute in the equation:

\(v + x\frac{dv}{dx} = v + \tan(v)\)

⇒ \(x\frac{dv}{dx} = \tan(v)\)

⇒ \(\frac{dv}{\tan(v)} = \frac{dx}{x}\)

⇒ \(\cot(v) dv = \frac{dx}{x}\)

Integrate both sides:

\(\int \cot(v) dv = \int \frac{dx}{x}\)

⇒ \(\ln|\sin(v)| = \ln|x| + \ln|C|\)

⇒ \(\ln|\sin(v)| = \ln|Cx|\)

Remove the logarithms:

⇒ \(\sin(v) = Cx\)

Substitute v = y/x:

⇒ \(\sin(\frac{y}{x}) = Cx\)

∴ The general solution is \(\sin(\frac{y}{x}) = Cx\).

Hence option 2 is correct

Calculation

\( \cfrac { dx }{ dy } =\cos { \left( x+y \right) } \)

Let \( x+y=v \implies \dfrac{dx}{dy}+1=\dfrac{dv}{dy} \)

⇒ \( \dfrac{dv}{dy}=1+\cos{v}=2\cos^2{\dfrac{v}{2}} \)

\(\Rightarrow \dfrac{dv}{\cos^2{\dfrac{v}{2}}}=2dy \)

⇒ \( \sec^2{\dfrac{v}{2}}dv=2dy \)

Integrating both sides:-

⇒ \( 2\tan{\dfrac{v}{2}}=2y+k \)

⇒ \( \tan{\left(\dfrac{x+y}{2}\right)}=y+c \)

Hence, option 1 is correct

Calculation

\(\dfrac {dy}{dx} = \tan \left (\dfrac {y}{x}\right ) + \left (\dfrac {y}{x}\right )\) ..... \((i)\)

Take, \(\dfrac {y}{x} = v\)

\(\implies y = vx\)

\(\implies \dfrac {dy}{dx} = v + x\dfrac {dv}{dx}\)

\(\therefore\) The given equation \((i)\) becomes

\(v + x\dfrac {dv}{dx} = \tan v + v\)

\(\implies \dfrac {1}{\tan v}dv = \dfrac {1}{x}dx\)

\(\implies \displaystyle \int \cot v\ dv = \int \dfrac {1}{x}dx\)

\(\implies \log |\sin v| = \log x + \log c=\log|xc|\)

\(\implies \sin v = xc\)

\(\therefore \sin \left (\dfrac {y}{x}\right ) = xc\)

Hence option 2 is correct

Calculation

\( \cfrac { dx }{ dy } =\cos { \left( x+y \right) } \)

Let \( x+y=v \implies \dfrac{dx}{dy}+1=\dfrac{dv}{dy} \)

⇒ \( \dfrac{dv}{dy}=1+\cos{v}=2\cos^2{\dfrac{v}{2}} \)

\(\Rightarrow \dfrac{dv}{\cos^2{\dfrac{v}{2}}}=2dy \)

⇒ \( \sec^2{\dfrac{v}{2}}dv=2dy \)

Integrating both sides:-

⇒ \( 2\tan{\dfrac{v}{2}}=2y+k \)

⇒ \( \tan{\left(\dfrac{x+y}{2}\right)}=y+c \)

Hence, option 1 is correct

Calculation

\(x~dy-y~dx=\sqrt{x^{2}+y^{2}}dx\)

⇒ \(\frac{dy}{dx} - \frac{y}{x} = \sqrt{1+(\frac{y}{x})^2}\)

Let \(y=vx\), then \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)

⇒ \(v + x\frac{dv}{dx} - v = \sqrt{1+v^2}\)

⇒ \(x\frac{dv}{dx} = \sqrt{1+v^2}\)

⇒ \(\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}\)

Integrating both sides:

⇒ \(\int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x}\)

⇒ \(\ln|v+\sqrt{1+v^2}| = \ln|x| + \ln C\)

⇒ \(v+\sqrt{1+v^2} = Cx\)

⇒ \(\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}} = Cx\)

⇒ \(y+\sqrt{x^2+y^2} = Cx^2\)

Given \(y(\sqrt{3})=1\), so

⇒ \(1+\sqrt{3+1} = C(\sqrt{3})^2\)

⇒ \(1+2 = 3C\)

⇒ \(C=1\)

∴ \(y+\sqrt{x^2+y^2} = x^2\)

Hence option 3 is correct

Concept:

Some useful formulas are:

\(\rm \int{ dx\over x}=logx+c\)

\(\rm \int{ dx \over {a^2+x^2}}={1\over a}tan^{-1}x+c\)

Calculation:

\(\rm x^2{ dy\over dx}= x^2+xy+y^2\)

⇒\(\rm { dy\over dx}= 1+{y\over x}+({y\over x})^2\)

Substituting y = vx and  \(\rm {{dy}\over {dx}} = v+x {{dv}\over{dx}} \)

⇒ \(\rm v+x{ dv\over dx}= 1+v+v^2\)

⇒ \(\rm x{ dv\over dx}= 1+v^2\)

Integrating both sides we get,

\(\rm \int{ dx\over x}=\int{ dv \over {1+v^2}}\)

⇒ \(\rm logx= tan^{-1}v+c\), c = constant of integration

Putting the value of v we get,

∴ \(\rm \log x= tan^{-1}{y\over x}+c\)

Concept:

Some useful formulas are:

\(\rm \int{{dx}\over {x}}=log x+c\)

\(\rm \int{cot\ x\ dx}=log(sin\ x)+c\)

Calculation:

\(\rm x {{dy}\over {dx}}=y+x\ tan {y\over x}\)

Substituting y=vx and \(\rm v+x {{dv}\over{dx}} = {{dy}\over {dx}}\)

\(\rm x [v+x{{dv}\over {dx}}]=vx+x\ tan {v}\)

\(\rm x [v+x{{dv}\over {dx}}]=x(v+ tan {v})\)

\(\rm \int{{dx}\over {x}}=\int{{dv}\over tan {v}}\)

\(\rm \int{{dx}\over {x}}=\int{cotv\ dv}\)

log x = log (sinv) + log c

x = c sinv

Putting the value of v we get,

\(\rm x = c\sin{y\over x}\)             

[Where c = constant of integration]

Concept:

If a differential equation has the form f(x,y)dy = g(x,y)dx then it is said to be a homogeneous differential equation if the degree of f(x,y) and g(x, y) is the same.

Some useful formulas are;

\(\rm \int {1\over x} dx =ln\ x + C\)

\(\rm \int {1\over 1+x^2} dx =tan^{-1} x + C\)

Calculation:

\(\rm {{dy}\over {dx}} = {{x+y}\over {x-y}}\)

Taking y = vx where \(\rm {{dy}\over{dx}} = v+ x{{dv}\over{dx}}\) then we get

\(\rm v+ x{{dv}\over{dx}} = {{x+vx}\over {x-vx}}\)

\(\rm v+ x{{dv}\over{dx}} = {{x(1+v)}\over {x(1-v)}}\)

\(\rm x{{dv}\over{dx}} = {{1+v}\over {1-v}} -v\)

\(\rm x{{dv}\over{dx}} = {{1+v-v+v^2}\over {1-v}} \)

\(\rm {{dx}\over{x}} = {{1-v}\over {1+v^2}} dv\)

Integrating the above equation we get,

\(\rm \int{{dx}\over{x}} = \int{{1}\over {1+v^2}}dv- \int{{v}\over {1+v^2}}dv\)

\(\rm \log x = \tan^{-1} v-{1\over 2}log(1+v^2)+c\)

Now finally putting the value of (v = y/x) in the equation we get,

\(\rm \log x = tan^{-1} {y\over x}-{1\over 2}log(1+({y\over x})^2)+c\)

Concept:

If a differential equation has the form f(x,y)dy = g(x,y)dx then it is said to be a homogeneous differential equation if the degree of f(x,y) and g(x, y) is the same.

Some useful formulas are;

\(\rm \int {1\over x} dx =ln\ x + C\)

\(\rm \int {1\over 1+x^2} dx =tan^{-1} x + C\)

Calculation:

\(\rm {{dy}\over {dx}} = {{x+y}\over {x-y}}\)

Taking y = vx where \(\rm {{dy}\over{dx}} = v+ x{{dv}\over{dx}}\) then we get

\(\rm v+ x{{dv}\over{dx}} = {{x+vx}\over {x-vx}}\)

\(\rm v+ x{{dv}\over{dx}} = {{x(1+v)}\over {x(1-v)}} \)

\(\rm x{{dv}\over{dx}} = {{1+v}\over {1-v}} -v\)

\(\rm x{{dv}\over{dx}} = {{1+v-v+v^2}\over {1-v}} \)

\(\rm {{dx}\over{x}} = {{1-v}\over {1+v^2}} dv\)

Integrating the above equation we get,

\(\rm \int{{dx}\over{x}} = \int{{1}\over {1+v^2}}dv- \int{{v}\over {1+v^2}}dv\)

\(\rm \log x = \tan v^{-1}-{1\over 2}log(1+v^2)+c\)

Now finally putting the value of y in the equation we get,

\(\rm \log x = tan^{-1} {y\over x}-{1\over 2}log(1+({y\over x})^2)+c\)

Concept:

Some useful formulas are:

\(\rm \int{ dx\over x}=logx+c\)

\(\rm \int{ dx \over {a^2+x^2}}={1\over a}tan^{-1}x+c\)

Calculation:

\(\rm x^2{ dy\over dx}= x^2+xy+y^2\)

⇒\(\rm { dy\over dx}= 1+{y\over x}+({y\over x})^2\)

Substituting y = vx and  \(\rm {{dy}\over {dx}} = v+x {{dv}\over{dx}} \)

⇒ \(\rm v+x{ dv\over dx}= 1+v+v^2\)

⇒ \(\rm x{ dv\over dx}= 1+v^2\)

Integrating both sides we get,

\(\rm \int{ dx\over x}=\int{ dv \over {1+v^2}}\)

⇒ \(\rm logx= tan^{-1}v+c\), c = constant of integration

Putting the value of v we get,

∴ \(\rm \log x= tan^{-1}{y\over x}+c\)

Concept:

Some useful formulas are:

\(\rm\int {dx\over x} =log x+c\)

\(\rm ∫ x^n dx = \frac {(x^{n+1})} {(n+1)} +C; \ n≠1\)

Calculation:

(xy + y2)dx = (x2 - xy)dy

Dividing both sides by x² we get,

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

Now substituting y = vx and \(\rm v+x {{dv}\over{dx}} = {{dy}\over {dx}}\)

\(\rm (v+v^2)dx=(1-v)dy\)

\(\rm {dy\over dx} ={{ (v+v^2)}\over{ (1-v)}}\)

\(\rm v+x {dv\over dx} ={{ (v+v^2)}\over{ (1-v)}}\)

\(\rm x {dv\over dx} ={{ (v+v^2-v+v^2)}\over{ (1-v)}}\)

\(\rm x {dv\over dx} ={{ 2v^2}\over{ (1-v)}}\)

Integrating the equation we get,

\(\rm\int {dx\over x} =\int{(1-v)dv\over {2v^2}}\)

\(\rm\int {dx\over x} =\int({{1\over {2v^2}}-{1\over {2v}}})dv\)

\(\rm \log x= {-1\over 2v}- {1\over 2}\log v+c\), c= constant of integration

Putting the value of v we get,

\(\rm \log x+ {x\over 2y}+{1\over 2}\ {log{y\over x}} = c\)

Concept:

Some useful formulas are:

\(\rm \int{{dx}\over {x}}=log x+c\)

\(\rm \int{cot\ x\ dx}=log(sin\ x)+c\)

Calculation:

\(\rm x {{dy}\over {dx}}=y+x\ tan {y\over x}\)

Substituting y=vx and \(\rm v+x {{dv}\over{dx}} = {{dy}\over {dx}}\)

\(\rm x [v+x{{dv}\over {dx}}]=vx+x\ tan {v}\)

\(\rm x [v+x{{dv}\over {dx}}]=x(v+ tan {v})\)

\(\rm \int{{dx}\over {x}}=\int{{dv}\over tan {v}}\)

\(\rm \int{{dx}\over {x}}=\int{cotv\ dv}\)

log x = log (sinv) + log c

x = c sinv

Putting the value of v we get,

\(\rm x = c\sin{y\over x}\)

[Where c = constant of integration]

Calculation:

x(x + y + z) = 9 ....(1)

y(x + y + z) = 16 ....(2)

z(x + y + z) = 144 .....(3)

By adding these three equations

x(x + y + z) + y(x + y + z) + z(x + y + z) = 9 + 16 + 144

⇒ (x + y + z)(x + y + z) = 169

⇒ (x + y + z)² = 169

⇒ (x + y + z) = 13

From eq. (1)

⇒ x(x + y + z) = 9

⇒ x(13) = 9

⇒ x = \(\frac{9}{13}\)

Concept:

Some useful formulas are:

\(\rm \int{ dx\over x}=logx+c\)

\(\rm \int{ dx \over {a^2+x^2}}={1\over a}tan^{-1}x+c\)

Calculation:

\(\rm x^2{ dy\over dx}= x^2+xy+y^2\)

⇒\(\rm { dy\over dx}= 1+{y\over x}+({y\over x})^2\)

Substituting y = vx and  \(\rm {{dy}\over {dx}} = v+x {{dv}\over{dx}} \)

⇒ \(\rm v+x{ dv\over dx}= 1+v+v^2\)

⇒ \(\rm x{ dv\over dx}= 1+v^2\)

Integrating both sides we get,

\(\rm \int{ dx\over x}=\int{ dv \over {1+v^2}}\)

⇒ \(\rm logx= tan^{-1}v+c\), c = constant of integration

Putting the value of v we get,

∴ \(\rm \log x= tan^{-1}{y\over x}+c\)

Given:

\(\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{f\left( {\frac{y}{x}} \right)}}{{f'\left( {\frac{y}{x}} \right)}}\)

Concept:

To solve such type of differential equations, simply put y = vx.

Calculation:

Putting y = vx

⇒ \(\frac{dy}{dx} = v + x\frac{dv}{dx} \) 

Now, \(\frac{{dy}}{{dx}} = \frac{y}{x} + \frac{{f\left( {\frac{y}{x}} \right)}}{{f'\left( {\frac{y}{x}} \right)}}\)

⇒ \( v + x\frac{dv}{dx} \) = v + \(\frac{f(v)}{f'(v)}\)

⇒ \(\frac{f'(v)}{f(v)}dv\) = \(\frac{dx}{x}\)

Integrating both sides -

⇒ ln|f(v)| = ln|x| + lnC

⇒ f(v) = Cx

putting v = y/x back,

⇒ f(y/x) = Cx