99. Prove that <math><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mi>c</mi><msup><mrow><mrow><mo>(</mo><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow></mrow><mrow><mn>2</mn></mrow></msup></mrow></math> is the general equation of the differential equation <math><mrow><mrow><mo>(</mo><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>−</mo><mn>3</mn><mi>x</mi><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mo>)</mo></mrow><mi>d</mi><mi>x</mi><mo>=</mo><mrow><mo>(</mo><mrow><msup><mrow><mi>y</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>−</mo><mn>3</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>y</mi></mrow><mo>)</mo></mrow><mi>d</mi><mi>y</mi></mrow></math> where c is a parameter.

⇒dydx=x3−3xy2y3−3x2y..........(1)This is a homogenous equation. To simplify it, we need to make the substitution as:y=vx⇒ddx(y)=ddx(vx)⇒dydx=v+xdvdxSubstituting the values of y and dvdx in equation (1), we get:v+xdvdx=x3−3x(vx)2(vx)3−3x2(vx)⇒v+xdvdx=1−3v2v3−3v⇒xdvdx=1−3v2v3−3v−v⇒xdvdx=1−3v2−v(v3−3v)v3−3v⇒xdvdx=1−v4v3−3v⇒(v3−3v1−v4)dv=dxxIntegrating both sides, we get:∫(v3−3v1−v4)dv=logx+logC'.........(2)Now,∫(v3−3v1−v4)dv=∫v3dv1−v4−3∫vdv1−v4⇒∫(v3−3v1−v4)dv=I1−3I2,Where,I1=∫v3dv1−v4andI2=∫vdv1−v4...........(3)Let,1−v4=t.∴ddv(1−v4)=dtdv⇒−4v3=dtdv⇒v3dv=−dt4Now,I1=∫−dt4=−logt=−14log(1−v4)And,I2=∫vdv1−v4=∫vdv1−(v2)2Let,v2=p.∴ddv(v2)=dpdv⇒2v=dpdv⇒vdv=p2⇒I2=12∫dp1−p2=12×2log|1+p1−p|=14log|1+v21−v2|Substituting the values of I1 and I2 in equation (3), we get:∫(v3−3v1−v4)dv=−14log(1−v4)−34log|1+v21−v2|Therefore, equation (2) becomes:14log(1−v4)−34log|1+v21−v2|=logx+logC'⇒−14log[(1−v4)(1+v21−v2)]=logC'x⇒(1+v2)4(1−v2)2=(C'x)−4⇒(1+y2x2)4(1−y2x2)2=1C'4x4⇒(x2+y2)4x4(x2−y2)2=1C'4x4⇒(x2−y2)2=C'4(x2+y2)4⇒(x2−y2)=C'2(x2+y2)2⇒x2−y2=C(x2+y2)2,whereC=C'2Hence, the given result is proved.

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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.

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Vishal Baghel | Contributor-Level 10

4 months ago

$\Rightarrow \frac{d y}{d x} = \frac{x^{3} - 3 x y^{2}}{y^{3} - 3 x^{2} y} . . . . . . . . . . (1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l} y = v x \\ \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (v x) \\ \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$

Substituting the values of y and $\frac{d v}{d x}$ in equation (1), we get:

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{x^{3} - 3 x {(v x)}^{2}}{{(v x)}^{3} - 3 x^{2} (v x)} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2} - v (v^{3} - 3 v)}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - v^{4}}{v^{3} - 3 v} \\ \Rightarrow (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \frac{d x}{x} \end{array}$

Integrating both sides, we get:

$\begin{array}{l} \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = l o g x + l o g C^{'} . . . . . . . . . (2) \\ N o w, \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \int \frac{v^{3} d v}{1 - v^{4}} - 3 \int \frac{v d v}{1 - v^{4}} \\ \Rightarrow \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = I_{1} - 3 I_{2}, W h e r e, I_{1} = \int \frac{v^{3} d v}{1 - v^{4}} a n d I_{2} = \int \frac{v d v}{1 - v^{4}} . . . . . . . . . . . (3) \end{array}$

$\begin{array}{l} L e t, 1 - v^{4} = t . \\ ∴ \frac{d}{d v} (1 - v^{4}) = \frac{d t}{d v} \\ \Rightarrow - 4 v^{3} = \frac{d t}{d v} \\ \Rightarrow v^{3} d v = - \frac{d t}{4} \\ N o w, I_{1} = \int - \frac{d t}{4} = - l o g t = - \frac{1}{4} l o g (1 - v^{4}) \end{array}$

$\begin{array}{l} A n d, I_{2} = \int \frac{v d v}{1 - v^{4}} = \int \frac{v d v}{1 - (v^{2})^{2}} \\ L e t, v^{2} = p . \\ ∴ \frac{d}{d v} (v^{2}) = \frac{d p}{d v} \\ \Rightarrow 2 v = \frac{d p}{d v} \\ \Rightarrow v d v = \frac{p}{2} \\ \Rightarrow I_{2} = \frac{1}{2} \int \frac{d p}{1 - p^{2}} = \frac{1}{2 \times 2} l o g | \frac{1 + p}{1 - p} | = \frac{1}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | \end{array}$

Substituting the values of $I_{1}$ and $I_{2}$ in equation (3), we get:

$\int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = - \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} |$

Therefore, equation (2) becomes:

$\begin{array}{l} \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | = l o g x + l o g C^{'} \\ \Rightarrow - \frac{1}{4} l o g [(1 - v^{4}) (\frac{1 + v^{2}}{1 - v^{2}})] = l o g C^{'} x \\ \Rightarrow \frac{{(1 + v^{2})}^{4}}{{(1 - v^{2})}^{2}} = {(C^{'} x)}^{- 4} \\ \Rightarrow \frac{{(1 + \frac{y^{2}}{x^{2}})}^{4}}{{(1 - \frac{y^{2}}{x^{2}})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow \frac{{(x^{2} + y^{2})}^{4}}{x^{4} {(x^{2} - y^{2})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow {(x^{2} - y^{2})}^{2} = C^{' 4} {(x^{2} + y^{2})}^{4} \\ \Rightarrow (x^{2} - y^{2}) = C^{' 2} {(x^{2} + y^{2})}^{2} \\ \Rightarrow x^{2} - y^{2} = C {(x^{2} + y^{2})}^{2}, w h e r e C = C^{' 2} \end{array}$

Hence, the given result is proved.

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99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.

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99. Prove that x2−y2=c(x2+y2)2 is the general equation of the differential equation (x3−3xy2)dx=(y3−3x2y)dy where c is a parameter.

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99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.