CBSE Class 12 Maths Chapter 9 Differential Equations Important Questions
Question 1.
- Write the order and degree of the differential equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
- Find the general solution of the differential equation \(\frac{dy}{dx}\) = a. where a is an arbitrary constant.
Solution:
1. Given differential equation is: \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
⇒ Order = 2 and degree is not defined since the given equation is not a polynomial in
derivatives
2. Given differential equation is \(\frac{dy}{dx}\) = a
⇒ dy = a dx
Integrating both sides; we get: ∫l dy =∫a dx ⇒ y = ax + C
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Question 2. Find the integrating factor of x \(\frac{dy}{dx}\) + (1 + x cot x) y = x
Solution:
x\(\frac{d y}{d x}+(1+x \cot x) y=x \Rightarrow \frac{d y}{d x}+y\left(\frac{1}{x}+\cot x\right)=1\)
Comparing it with \(\frac{dy}{dx}\) + Py = Q; we get:
P = \(\left(\frac{1}{x}+\cot x\right)\) and Q=1
⇒ \(I \cdot F .=e^{\int P d x}=e^{\int\left(\frac{1}{x}+\cot x\right) d x}=e^{(\mathrm{log} x+\log \sin x)}\)
or I.F. = \(e^{\log (x \sin x)}=x \sin x\)
Class 12 Maths Differential Equations Solved Examples
Question 3. Find the general solution of the differential equation: \(\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^x}\)
Solution:
Given differential equation is \(\frac{d y}{d x}=\frac{3 e^{2 x}+3 e^{4 x}}{e^x+e^{-x}}\)
⇒ \(\frac{d y}{d x}=\frac{3 e^{2 x}\left(1+e^{2 x}\right)}{e^{-x}\left(1+e^{2 x}\right)} \Rightarrow \frac{d y}{d x}=3 e^{3 x}\)
⇒ \(\int 1 \cdot d y=\int 3 e^{3 x} d x \Rightarrow y=3\left(\frac{e^{3 x}}{3}\right)+C \Rightarrow y=e^{3 x}+C\)
Question 4. Find the general solution of the differential equation: log(\(\frac{dy}{dx}\)) = ax + by.
Solution:
Given that, log(\(\frac{dy}{dx}\)) = ax + by.
⇒ \(\frac{d y}{d x}=e^{a x+b y}=e^{a x} \cdot e^{b y} \Rightarrow \frac{d y}{e^{b y}}=e^{a x} d x\)
On integrating both sides; we get:
⇒ \(\int e^{-b y} d y=\int e^{a x} d x \Rightarrow \frac{e^{-b y}}{-b}=\frac{e^{a x}}{a}+c_1 \)
⇒ \(e^{-b y}=-\frac{b}{a} e^{a x}+C \text { where } C=-b c_1\)
Question 5. Find the general solution of the differential equation: edy/dx = x²
Solution:
Given that edy/dx = x²
Taking log on both sides; we get:
⇒ log edy/dx = log x²
⇒ \(\frac{dy}{dx}\) = 2 logx
⇒ dy = 2logxdx
On integrating both sides; we have:
⇒ \(\int 1 \cdot d y=2 \int 1 \cdot \log x d x \Rightarrow y=2\left[\log x \int 1 \cdot d x-\int\left(\frac{1}{x} \cdot \int 1 d x\right) d x\right]\)
⇒ y = 2x logx -2x+c
Question 6. If the solution of the differential equation \(\frac{d y}{d x}=\frac{2 x y-y^2}{2 x^2} \text { is } \frac{a x}{y}=b \log |x|+C \text {; }\) then find the value of ‘a’ and ‘b’.
Solution:
Given, \(\frac{a x}{y}=b \log |x|+C\)
Differentiating with respect to x, \(a\left[\frac{y-x \frac{d y}{d x}}{y^2}\right]=\frac{b}{x}\)
⇒ \(y-x \frac{d y}{d x}=\frac{b y^2}{a x} \Rightarrow x \frac{d y}{d x}=y-\frac{b y^2}{a x} \)
⇒ \(\frac{d y}{d x}=\frac{a x y-b y^2}{a x^2}\) will be the required differential equation
Now comparing with \(\frac{d y}{d x}=\frac{2 x y-y^2}{2 x^2}\), we get a = 2 and b = 1
First-Order Differential Equations Problems Class 12 Maths
Question 7. Find the particular solution of the differential equation x \(\frac{dy}{dx}\) – y = x².ex. given y(1) = 0.
Or,
Find the general solution of the differential equation x \(\frac{dy}{dx}\) = y(log y – log x +1).
Solution:
Given differential equation is x \(\frac{dy}{dx}\) – y = x² · ex
⇒ \(\frac{dy}{dx}\) — \(\frac{y}{x}\)= x · ex
This is a linear differential equation of the form \(\frac{dy}{dx}\) + Py = Q
where P = –\(\frac{1}{x}\) and Q = x · ex
∴ I.F. = \(\mathrm{e}^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}\)
Hence, the general solution is given by \(y\left(\frac{1}{x}\right)=\int x \cdot e^x\left(\frac{1}{x}\right) d x\)
⇒ \(\frac{y}{x}=\int e^x d x \Rightarrow \frac{y}{x}=e^x+c\)
Substituting x = 1 and y = 0, we get; 0 = e1 + c
⇒ c = -e
∴ Required particular solution: \(\frac{y}{x}\)= ex – e
Or,
Given differential equation is \(x \frac{d y}{d x}=y(\log y-\log x+1) \Rightarrow \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)\)
Put y = \(v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \Rightarrow v+x \frac{d v}{d x}=v(\log v+1)\)
⇒ \(\frac{1}{v \log v} d v=\frac{d x}{x}\)
Integrating both sides: \(\int \frac{1}{v \log v} d v=\int \frac{1}{x} d x\)
⇒ log(log v) = log x + log c = log cx ⇒ log v = cx
⇒ log \(\frac{y}{x}\) = cx, which is the required general solution.
Question 8. Solve the Differential equation \(\left(1+e^{y / x}\right) d y+e^{y / x}\left(1-\frac{y}{x}\right) d x=0(x \neq 0)\)
Solution:
Given differential equation is \(\left(1+e^{y / x}\right) d y+e^{y / x}\left(1-\frac{y}{x}\right) d x=0(x \neq 0)\)
⇒ \(\left(1+e^{y / x}\right) d y=\left(\frac{y}{x}-1\right) e^{y x} d x \Rightarrow \frac{d y}{d x}=\frac{e^{y x}\left(\frac{y}{x}-1\right)}{1+e^{y x}}\)
Put \(\frac{y}{x}=v \text { i.e. } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
∴ \(v+x \frac{d v}{d x}=\frac{e^v(v-1)}{1+e^v} \Rightarrow x \frac{d v}{d x}=\frac{e^v(v-1)}{1+e^v}-v\)
⇒ \(x \frac{d v}{d x}=\frac{v \cdot e^v-e^v-v-v e^v}{1+e^v} \Rightarrow x \frac{d v}{d x}=-\frac{\left(e^v+v\right)}{1+e^v}\)
⇒ \(\int \frac{1+e^v}{v+e^v} d v=-\int \frac{d x}{x} \Rightarrow \log \left|v+e^v\right|=-\log |x|+\log C\)
⇒ \(\log \left|v+e^v\right|=\log \left|\frac{C}{x}\right| \Rightarrow v+e^v=\frac{C}{x}\)
⇒ \(x\left(\frac{y}{x}+e^{y / x}\right)=C \Rightarrow y+x e^{y / x}=C\)
Question 9. Solve the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\)
Or,
Solve the differential equation \(\frac{d y}{d x}=-\left[\frac{x+y \cos x}{1+\sin x}\right]\)
Solution:
Given differential equation is: \(x \frac{d y}{d x}=y-x \tan (y / x) \Rightarrow \frac{d y}{d x}=\left(\frac{y}{x}\right)-\tan \left(\frac{y}{x}\right)\)
The given equation is a homogenous differential equation.
Put \(\frac{y}{x}=v \text { i.e. } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
∴ \(v+x \frac{d v}{d x}=v-\tan v \Rightarrow x \frac{d v}{d x}=-\tan v \Rightarrow \frac{d v}{\tan v}=-\frac{d x}{x}\)
Integrating both sides; we get:
⇒ \(\int \frac{d v}{\tan v}=\int \frac{-d x}{x}\)
⇒ \(\int \cot v d v=-\int \frac{1}{x} d x \Rightarrow \log |\sin v|=-\log |x|+\log C\)
⇒ \(\log |\sin v|=\log \left|\frac{C}{x}\right|\)
⇒ \(\sin \left(\frac{y}{x}\right)=C / x \Rightarrow x \sin \left(\frac{y}{x}\right)=C\)
Or,
Given differential equation is: \(\frac{d y}{d x}=-\left[\frac{x+y \cos x}{1+\sin x}\right] \Rightarrow \frac{d y}{d x}+\left(\frac{\cos x}{1+\sin x}\right) y=\frac{-x}{1+\sin x}\)…..(1)
Equation(1) is a linear differential equation of the form \(\frac{dy}{dx}\)+ Py = Q;
where, P = \(\left(\frac{\cos x}{1+\sin x}\right)\) and \(Q=\left(\frac{-x}{1+\sin x}\right)\)
⇒ I.F. =\(\mathrm{e}^{\int^{P d x}}=\mathrm{e}^{\int \frac{\cos x}{1+\sin x} d x}=e^{\mathrm{sin}|+\sin x|}=1+\sin x\)
Hence, the solution is y(I.F) = ∫(Q.I.F)dx + C
⇒ \(y(1+\sin x)=\int\left(\frac{-x}{1+\sin x}\right) \times(1+\sin x) d x+C\)
⇒ \(y(1+\sin x)=\int-x d x+C \quad \text { or } y(1+\sin x)=\frac{-x^2}{2}+C\)
Differential Equations Important formulas For Class 12 CBSE
Question 10. Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec²ydy = 0, given that y = π/4 when x = 0.
Or,
Find the particular solution of the differential equation \(\frac{dy}{dx}\) + 2ytanx = sin x, given that y = 0 when x = π/4.
Solution:
The given differential equation is ex tan ydx + (2- ex)sec² ydy = 0
∴ \(\left(2-e^x\right) \sec ^2 y d y=-e^x \tan y d x \Rightarrow \frac{\sec ^2 y}{\tan y} d y=\frac{-e^x}{2-e^x} d x\)
Integrating both sides, we get \(\int \frac{\sec ^2 y}{\tan y} d y=\int \frac{-e^x}{2-e^x} d x\)
⇒ log(tany) = log(2-ex) + logC ⇒ log (tan y) = log[C(2 — ex)]
⇒ tan y = C (2 – ex)
y = π/4 when x = 0 ⇒ C = 1
So, the required particular solution is tan y = 2-ex
The given differential equation is \(\frac{dy}{dx}\) + 2y tax = sin x
This is a linear equation of the form \(\frac{dy}{dx}\) + Py = Q
where P = 2 tan x and Q = sin x
Now, I.F. \(=\mathrm{e}^{\int \mathrm{Pdx} x}=\mathrm{e}^{\int 2 \mathrm{tan} x \mathrm{xdx}}=\mathrm{e}^{2 \log \sec x}=\mathrm{e}^{\log \sec ^2 x}=\sec ^2 \mathrm{x}\)
The general solution of the given differential equation is y (I.F) = ∫(Q x I.F) dx + C
⇒ y(sec²x)= ∫(sin x.sec²x)dx + C
⇒ ysec²x = ∫(sec x.tanx)dx + C
⇒ y sec² x = sec x + C…..(1)
Now, y = 0 when x = π/3
Therefore, 0 x sec² π/3 = sec π/3 + C ⇒ 0 = 2 + C ⇒ C = -2
Substituting C = -2 in equation (1). we get: ysec²x – sec x – 2 ⇒ y = cos x – 2cos²x
Hence, the required solution of the given differential equation is y = cos x – 2cos²x.