Class 12 NCERT Solutions- Mathematics Part ii – Chapter 9– Differential Equations Exercise 9.4
Last Updated : 17 May, 2024
NCERT solutions Class 12 Chapter 9 Exercise 9.4 consists of 16 questions that impart a clear understanding of Differential Equations. Learn about the concept used and the solution to Chapter 9– Integrals Exercise 9.4 in this article.
Question 1. dy/dx = (x2 + y2) / (x2 + xy)
Solution:
Given equation:
\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}
F(x,y)=\frac{x^2+y^2}{x^2+xy} \\F(λx,λy)\frac{(λx)^2+(λy^2)}{(λx^2)+(λx)(λy)}\\=\frac{x^2+y^2}{x^2+xy}\\=λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx
\frac{dy}{dx}=v+x\frac{dv}{dx} \\v+x\frac{dv}{dx}=\frac{x^2+(vx)^2}{x^2+x(vx)} \\v+x\frac{dv}{dx}=\frac{1+v^2}{1+v} \\x\frac{dv}{dx}=\frac{1+v^2}{1+v} \\v=log[\frac{C}{x(1-v)^2}] \\\frac{y}{x}=log[\frac{C}{x(1-\frac{y}x)^2}] \\\frac{Cx}{(x-y)^2}=e^\frac{y}{x}
(x - y)2 = Cxey/x
Question 2. y' = \frac{x+y}{x}
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{x+y}{x}\\F(x,y)=\frac{x+y}{x} \\F(λx,λy)=\frac{λx+λy}{λx}\\=\frac{x+y}{x}\\=λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=v+x\frac{dv}{dx} \\v+x\frac{dv}{dx}=\frac{x+vx}{x} \\v+x\frac{dv}{dx}=1+v \\\frac{dv}{dx}=\frac{1}{x} \\dv=\frac{dx}{x}
y = xlog|x| + Cx
Question 3. (x-y)dy - (x+y)dx = 0
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{x+y}{x-y}\\F(x,y)=\frac{x+y}{x-y} \\F(λx,λy)=\frac{λx+λy}{λx-λy}\\=\frac{x+y}{x-y}\\=λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=\frac{d}{dx}(vx) \\v+x\frac{dv}{dx}=\frac{1+v}{1-v} \\x\frac{dv}{dx}=\frac{1+v^2}{1-v} \\tan^{-1}[\frac{y}x]=\frac{1}2log(x^2+y^2)+C
tan-1(y/x) = 1/2[log(x2 + y2)] + C
Question 4. (x2-y2)dx + 2xy dy = 0
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=-\frac{(x^2-y^2)}{2xy}\\ ⇒F(x,y)=-\frac{(x^2-y^2)}{2xy}\\ ⇒F(λx,λy)=-\frac{((λx)^2-(λy)^2)}{2(λx)(λy)}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
⇒\frac{dy}{dx}=\frac{d}{dx}(vx)\\ ⇒\frac{dy}{dx}=v+x\frac{dv}{dx}\\ ⇒v+x\frac{dv}{dx}=-[\frac{x^2-(vx)^2}{2x.(vx)}]\\ ⇒v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\\ ⇒x\frac{dv}{dx}=\frac{v^2-1}{2v}-v\\ ⇒\frac{2v}{1+v^2}dv=-\frac{dx}{x}
⇒ x2 + y2 = Cx
Question 5. x^2\frac{dy}{dx}=x^2-2y^2+xy
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{x^2-2y^2+xy}{x^2}\\ ⇒F(x,y)=\frac{x^2-2y^2+xy}{x^2}\\ ⇒F(λx,λy)=\frac{(λx)^2-2(λy)^2+λxλy}{(λx)^2}\\ =\frac{x^2-2y^2+xy}{x^2}\\=λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
⇒\frac{dy}{dx}=v+x\frac{dv}{dx}\\ ⇒v+x\frac{dv}{dx}=\frac{x^2-2(vx)^2+x(vx)}{x^2}\\ ⇒x\frac{dv}{dx}=1-2v^2\\ ⇒\frac{dv}{1-2v^2}=\frac{dx}{x}\\ ⇒\frac{1}2[\frac{dv}{[\frac{1}{\sqrt2}]^2-v^2}]=\frac{dx}{x}\\ ⇒\frac{1}{2\sqrt{2}}log|\frac{x+\sqrt2y}{x-\sqrt2y}|=log|x|+C
Question 6. x dy - y dx = \sqrt{x^2+y^2}dx
Solution:
Given equation can be rearranged as:
x dy - y dx = \sqrt{x^2+y^2}dx\\ ⇒xdy=[y+\sqrt{x^2+y^2}]dx\\ ⇒\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}x\\ ⇒F(x,y)=\frac{y+\sqrt{x^2+y^2}}x\\ ⇒F(λx,λy)=\frac{λy+\sqrt{(λx)^2+(λy)^2}}{λx}\\ =\frac{y+\sqrt{x^2+y^2}}x\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=v+x\frac{dv}{dx}\\ ⇒v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^2+(vx)^2}}x\\ ⇒v+x\frac{dv}{dx}=v+\sqrt{1+v^2}\\ ⇒\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}\\ ⇒log|v+\sqrt{1+v^2}|=log|x|+logC\\ ⇒log|\frac{y}x+\sqrt{1+\frac{y^2}{x^2}}|=log|Cx|\\ ⇒y+\sqrt{x^2+y^2}=Cx^2
Question 7. [xcos(\frac{y}x)+ysin(\frac{y}x)]ydx=[ysin(\frac{y}x)-xcos(\frac{y}x)]xdy
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{[xcos(\frac{y}x)+ysin(\frac{y}x)]y}{[ysin(\frac{y}x)-xcos(\frac{y}x)]x}\\ ⇒F(x,y)=\frac{[xcos(\frac{y}x)+ysin(\frac{y}x)]y}{[ysin(\frac{y}x)-xcos(\frac{y}x)]x}\\ ⇒F(λx,λy)=\frac{[λxcos(\frac{λy}λx)+λysin(\frac{λy}λx)]λy}{[λysin(\frac{λy}λx)-λxcos(\frac{λy}λx)]λx}\\ ={[xcos(\frac{y}x)+ysin(\frac{y}x)]y}{[ysin(\frac{y}x)-xcos(\frac{y}x)]x}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=v+x\frac{dv}{dx}\\ ⇒v+x\frac{dv}{dx}=\frac{(xcosv+vxsinv)vx}{(vxsinv-xcosv)x}\\ ⇒x\frac{dv}{dx}=\frac{vcosv+v^2sinv}{vsinv-cosv}-v\\ ⇒x\frac{dv}{dx}=\frac{2vcosv}{vsinv-cosv}\\ ⇒[\frac{secv}{v}]=Cx^2\\ ⇒sec[\frac{y}x]=Cxy\\ ⇒xycos[\frac{y}x]=k
Question 8. x\frac{dy}{dx}-y+xsin[\frac{y}{x}]
Solution:
Given equation can be rearranged as:
x\frac{dy}{dx}=y-xsin[\frac{y}x]\\ ⇒\frac{dy}{dx}=\frac{y-xsin[\frac{y}x]}{x}\\ ⇒F(x,y)=\frac{y-xsin[\frac{y}x]}{x}\\ ⇒F(λx,λy)=\frac{λy-λxsin[\frac{λy}λx]}{λx}\\ =\frac{y-xsin[\frac{y}x]}{x}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
⇒\frac{dy}{dx}=v+x\frac{dv}{dx}\\ ⇒v+x\frac{dv}{dx}=\frac{vx-xsinx}{x}\\ ⇒v+x\frac{dv}{dx}=v-sinv\\ ⇒\frac{dv}{sinv}=-\frac{dx}{x}\\ ⇒cosecvdv=-\frac{dx}{x}\\ ⇒log|cosecvdv-cotv|=-logx+logC=log\frac{C}x\\ ⇒cosec[\frac{y}x]-cot[\frac{y}x]=\frac{C}x\\ ⇒x[1-cos[\frac{y}x]]=Csin\frac{y}x
Question 9. y dx + x log[\frac{y}{x}]dy-2x dy = 0
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{y}{2x-xlog[\frac{y}x]}\\ ⇒F(x,y)=\frac{y}{2x-xlog[\frac{y}x]}\\ ⇒F(λx,λy)=\frac{λy}{2λx-λxlog[\frac{λy}λx]}\\ =\frac{y}{2x-xlog[\frac{y}x]}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
⇒\frac{dy}{dx}=v+x\frac{dv}{dx}\\ ⇒v+x\frac{dv}{dx}=\frac{vx}{2x-xlogv}\\ ⇒x\frac{dv}{dx}=\frac{v}{2-logv}-v\\ ⇒x\frac{dv}{dx}=\frac{vlogv-v}{2-logv}\\ log[\frac{y}{x}]-1=Cy
Question 10. [1+e^{\frac{y}x}]dx+e^{\frac{x}y}[1-\frac{x}y]dy=0
Solution:
Given equation can be rearranged as:
\frac{dx}{dy}=\frac{-e^{\frac{x}y}[1-\frac{x}y]}{1+e^{\frac{x}y}}\\ ⇒F(x,y)=\frac{-e^{\frac{x}y}[1-\frac{x}y]}{1+e^{\frac{x}y}}\\ ⇒F(λx,λy)=\frac{-e^{\frac{λx}{λy}}[1-\frac{λx}{λy}]}{1+e^{\frac{λx}{λy}}}\\ =\frac{-e^{\frac{x}y}[1-\frac{x}y]}{1+e^{\frac{x}y}}\\ =λ^0F(λx,λy)
Thus, the given equation is homogenous.
Let y = vx.
⇒\frac{dx}{dy}=v+y\frac{dv}{dy}\\ ⇒v+y\frac{dv}{dy}=\frac{-e^y(1-y)}{1+e^v}\\ ⇒y\frac{dv}{dy}=-[\frac{v+e^v}{1+e^v}]\\ ⇒[\frac{v+e^v}{1+e^v}]dv=-\frac{dy}{y}\\ ⇒log[v+e^v]=log[\frac{c}y]\\ ⇒x+ye^{\frac{y}x}=C
For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:
Question 11. (x + y)dy + (x – y)dx = 0; y = 1 when x = 1
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{-(x-y)}{x+y}\\F(x,y)=\frac{-(x-y)}{x+y} \\F(λx,λy)=\frac{-λx+λy}{λx+λy}\\=\frac{-(x-y)}{x+y}\\=λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=\frac{d}{dx}(vx) \\v+x\frac{dv}{dx}=\frac{v-1}{v+1} \\x\frac{dv}{dx}=\frac{v-1}{v+1}-v \\x\frac{dv}{dx}=\frac{-(1+v^2)}{v+1} \\-\frac{dx}x=[\frac{v}{1+v^2}+\frac{1}{1+v^2}]\\ log(x^2+y^2)+2tan^{-1}\frac{y}x=2k
Since y = 1 when x = 1, we have:
⇒ log 2 + 2 tan-11 = 2k
⇒ π/2 + log 2 = 2k
Thus, log(x^2+y^2)+2tan^{-1}\frac{y}x=\frac{\pi}2+log2 is the required equation.
Question 12. x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{-(xy+y^2)}{x^2}\\ F(x,y)=\frac{-(xy+y^2)}{x^2} \\F(λx,λy)=\frac{λxλy+(λy)^2}{(λx)^2}\\ =\frac{-(xy+y^2)}{x^2}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=\frac{d}{dx}(vx) \\v+x\frac{dv}{dx}=\frac{-(x.vx+(vx)^2)}{x^2} \\x\frac{dv}{dx}=-v(v+2) \\x\frac{dv}{dx}=\frac{-(1+v^2)}{v+1} \\-\frac{dx}x=\frac{dv}{v(v+2)}\\ \frac{1}2[\frac{2}{v(v+2)}]dv=-\frac{dx}x\\ \frac{x^2y}{y+2x}=C^2
Since y = 1 when x = 1, we have:
C2 = 1/3
Thus, y + 2x = 3x2y is the required equation.
Question 13. [xsin^2(\frac{x}y-y)]dx+xdy=0; y = \frac{\pi}4\space when\space x = 1
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{-[xsin^2(\frac{x}y-y)]}{x}\\ F(x,y)=\frac{-[xsin^2(\frac{x}y-y)]}{x}\\ \\F(λx,λy)=\frac{-[λxsin^2(\frac{λx}λy-λy)]}{λx}\\ =\frac{-[xsin^2(\frac{x}y-y)]}{x}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=\frac{d}{dx}(vx) \\v+x\frac{dv}{dx}=\frac{-(xsin^2v-vx)}{x} \\x\frac{dv}{dx}=v-sin^2v \\x\frac{dv}{dx}=-sin^2v \\-\frac{dv}{sin^2v}=\frac{dx}x\\ cot(\frac{y}x)=log|Cx|
Now, y = π/4 when x = 1, we have:
1 = log C or C = e.
Thus, cot(\frac{y}x)=log|ex| is the required equation.
Question 14. \frac{dy}{dx}-\frac{y}{x}+cosec(\frac{y}x)=0; y = 0 when x = 1.
Solution:
Given equation can be rearranged as:
\frac{dy}{dx}=\frac{y}{x}-cosec(\frac{y}{x})\\ F(x,y)=\frac{y}{x}-cosec(\frac{y}{x})\\ \\F(λx,λy)=\frac{λy}{λx}-cosec(\frac{λy}λ{x})\\ =\frac{y}{x}-cosec(\frac{y}{x})\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=\frac{d}{dx}(vx) \\v+x\frac{dv}{dx}=v-cosec\space v \\-\frac{dv}{cosec\space v}=\frac{dx}{x}\\ -sin\space vdv=\frac{dx}{x}\\ cos(\frac{y}x)=log|Cx|
Now, y = 0 when x = 1, we have:
cos(0) = log C
⇒ C = e
Thus, cos(\frac{y}x)=log|ex| is the required solution.
Question 15. 2xy+y^2-2x^2\frac{dy}{dx}=0; y = 2 when x = 1
Solution:
Given equation can be rearranged as:
2x^2\frac{dy}{dx}=2xy+y^2\\ \frac{dy}{dx}=\frac{2xy+y^2}{2x^2}\\ F(x,y)=\frac{2xy+y^2}{2x^2}\\ \\F(λx,λy)=\frac{2λxλy+(λy)^2}{2(λx)^2}\\ =\frac{2xy+y^2}{2x^2}\\ =λ^0F(x,y)
Thus, the given equation is homogenous.
Let y = vx.
\frac{dy}{dx}=\frac{d}{dx}(vx) \\v+x\frac{dv}{dx}=\frac{2x(vx)+(vx)^2)}{2x^2} \\v+x\frac{dv}{dx}=v+\frac{v^2}{2}\\ -\frac{2x}y=log|x|+C
Now, y = 2 when x = 1, we have:
-1 = log(1) + C
C = -1
Thus, y=\frac{2x}{1-log|x|} is the required equation.
Question 16. A homogeneous differential equation of the from can be solved by making the substitution.
- (A) y = vx
- (B) v = yx
- (C) x = vy
- (D) x = v
Solution:
Option (C) is Correct
Question 17. Which of the following is a homogeneous differential equation?
- (A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
- (B) (xy) dx – (x3+ y3) dy = 0
- (C) (x3+ 2y2) dx + 2xy dy = 0
- (D) y2 dx + (x2– xy – y2) dy = 0
Solution:
Option (D) is Correct