Alternative English Year - 2017 Full Marks : 100
Class 12 Ncert Solutions Maths Differential Equations
Exercise 9.1
Next Exercises
Exercise 9.1 &9.2
Exercise 9.4
Exercise 9.6
Determine order and degree(if defined) of differential equations given in Exercises 1 to 10.
1. \(\frac{d^4y}{dx^4}+ sin(y''')\)
Ans: The order of the differential equation is 4 because the highest power of the highest derivative is 4. The degree of the given equation is not defined since it is nota polynomial equation
2. \(y^{'} + 5y = 0\)
Ans: The order of the equation is 1 and the degree of the equaion is also 1.
3. \(\left(\frac{ds}{dt}\right)^4 + 3s\frac{d^2s}{dt^2} = 0\)
Ans: The highest order derivative present in the given differential equation is \(\frac{d^2s}{dt^2}\), so its order is 2. It is a polynomial equation in \(\frac{d^2s}{dt^2}\) and \(\left(\frac{ds}{dt}\right)^4 \) and the highest power raised to \(\frac{d^2s}{dt^2}\) is 1, so its degree is 1.
4. \(\left(\frac{d^2y}{dx^2}\right)^2 + cos\left(\frac{dy}{dx}\right)=0\)
Ans: The highest order derivative present in the given differential equation is \(\frac{d^2y}{dx^2}\), so its order is 2. It is not polynomial equation so its degree is not defined.
5. \(\frac{d^2y}{dx^2}=cos3x + sin3x\)
Ans: The highest order derivative present in the given differential equation is \(\frac{d^2s}{dt^2}\), so its order is 2. It is a polynomial equation in \(\frac{d^2y}{dx^2}\) and the highest power raised to \(\frac{d^2x}{dy^2}\) is 1, so its degree is 1.
6. \((y^{'''})^2 + (y^{''})^2 + (y^{'})^4 + y^5 = 0\)
Ans: The highest order derivative present in the given differential equation is \(y^{'''}\), so its order is 3. It is a polynomial equation in \(y^{'''}\) , \(y^{''}\) and \(y^{'}\) and the highest power raised to \(y^{'''}\) is 2, so its degree is 2.
7. \(y^{'''} + 2y^{''} + y^{'} = 0\)
Ans: The highest order derivative present in the given differential equation is \(y^{'''}\), so its order is 3. It is a polynomial equation in \(y^{'''}\) , \(y^{''}\) and \(y^{'}\) and the highest power raised to \(y^{'''}\) is 1, so its degree is 1.
8. \(y{'} + y = e^x\)
Ans: The highest order derivative present in the given differential equation is \(y^{'}\), so its order is 1. It is a polynomial equation in \(y^{'}\) and the highest power raised to \(y^{'}\) is 1, so its degree is 1.
9. \(y^{''} + (y^{'})^2 +2y = 0\)
Ans: The highest order derivative present in the given differential equation is \(y^{''}\), so its order is 2. It is a polynomial equation in \(y^{''}\) and \(y^{'}\) and the highest power raised to \(y^{''}\) is 2, so its degree is 1.
10. \(y^{''} + 2y^{'} + siny = 0\)
Ans: The highest order derivative present in the given differential equation is \(y^{''}\), so its order is 2. It is a polynomial equation in \(y^{''}\) and \(y^{'}\) and the highest power raised to \(y^{''}\) is 2, so its degree is 1.
11. The degree of the differential equation
\(\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + sin\left(\frac{dy}{dx}\right) + 1=0\) is
(A) 3 (B) 2 (C) 1 (D) not defined
Ans: It is not a polynomial degree so the degree is not defined. The correct option is (D)
12. The order of the differential equation
\(2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} +y = 0 \) is
(A) 2 (B) 1 (C) 0 (D) not defined
Ans: The highest order derivative present in the given differential equation is \(\frac{d^2y}{dx}\), so its order is 2. The correct option is (a) 2.
Exercise 9.2
In each of the Exercises 1 to 10 verify that the given functions(explicit or implicit) is a solution of the corresponding differential equation:
1. \(y =e^x + 1\) : \(y^{''}- y^{'}= 0\)
Sol: Given function is \(y =e^x + 1\) . Differentiating both sides of the equation w.r.t x, we get
$$\frac{dy}{dx}=e^x\tag{1}$$
Now differentiating (1) with respect to x, we have
$$\frac{d^2y}{dx^2}= e^x\tag{2}$$
Substituting th vaules of \(\frac{d^2y}{dx^2},\frac{dy}{dx}\) in the given differential equation, we get
L.H.S = \(y^{''} -y^{'}=e^x - e^x=0\) = R.H.S
Therefore, the given function is a solution of the given differential equation.
2. \(y = x^2 + 2x +C\) : \(y^{'}- 2x -2= 0\)
Sol: Given function is \(y = x^2 + 2x +C\) . Differentiating both sides of equation w.r.t x, we get
$$\frac{dy}{dx}=y^{'}= 2x + 2 \tag{1}$$
Substituting this value in the given differential equation we get
L.H.S = \(y^{'}- 2x -2= 2x-2 -(2x-2) =0\) = R.H.S
Therefore the given function is a solution of the given differential equation.
3. y = cosx + C : \(y^{'}+ \sin x= 0\)
Sol: Given function is
$$y = cosx + C \tag{1}$$
Differentiating both sides of the equation (1) w.r.t x we get
$$ \frac{dy}{dx} = y^{'} = -\sin x$$
Substituting the value of \(y^{'}\) in given differential equation we get
L.H.S = \(-\sin x +\sin x= 0 \) = R.H.S
Therefore, the given function is a solution of the given differential equation.
4. \(y=\sqrt{1+x^2}\) : \(y^{'} =\frac{xy}{1+x^2}\)
Sol: the given funtion is
$$y=\sqrt{1+x^2} \tag{1}$$
Differentiating both sides of eq(1) with respect to x we get
$$\frac{dy}{dx} =y^{'} =\frac{2x}{2\sqrt{1+ x^2}}=\frac{x}{\sqrt{1+x^2}}$$
Substituting the value of \( y^{'}\) in the given differential equation we get
L.H.S = \( y^{'}=\frac{x}{1+x^2}\)
R.H.S = \(\frac{xy}{1+x^2}=\frac{x\sqrt{1+x^2}}{1+x^2}=\frac{x}{\sqrt{1+x^2}}\) = L.H.S
Therefore, the given function is a solution of the given differentail equation.
5. y = Ax : \(xy^{'}= y\)
Sol: Given function is y = Ax
Differentiating both sides of this equation we get
$$ \frac{dy}{dx} = y^{'} = A$$
Substituting the value of \(y^{'}\) in the given differential equation we get
L.H.S = \(xy^{'}=xA =y\) = R.H.S
Therefore the given function is a solution of the given differential equation.
6. \(y=x\sin x\) : \(xy^{'} = y + x\sqrt{x^2 - y^2}(x\neq0 \ and \ x>y \ or\ x<-y)\)
Sol: Given function is
$$ y =\sin x\tag{1}$$
Differentiating eq(1) with respect to x we get
$$ \frac{dy}{dx} = y^{'} =\sin x + x\cos x$$
Substituting the value of \(y^{'}\) in the given differential equation we get
L.H.S = \(xy^{'}=x(\sin x+x\cos x)=x\sin x+x^2\cos x\)
R.H.S = \(y + x\sqrt{x^2 - y^2}= x\sin x + x\sqrt{x^2 - x^2\sin ^2x}= x\sin x + x\sqrt{x^2\cos^2 x}=x\sin x+x^2\cos x\)
\(\therefore\) L.H.S = R.H.S
So the given function is asolution of the given differential function.
7. xy = \(\log\)y + C : \(y^{'} =\frac{y^2}{1-xy}(xy\neq1)\)
Sol: Given function is
$$xy = \log y + C \tag{1}$$
Differentiating both sides of eq(1) with respect to x we get
\begin{align}
x\frac{dy}{dx} + y &=\frac{1}{y}\frac{dy}{dx}\\
x\frac{dy}{dx} - \frac{1}{y}\frac{dy}{dx} &=- y\\
\frac{dy}{dx}(x-\frac{1}{y}) &=-y\\
\frac{dy}{dx}\frac{xy-1}{y}&=-y\\
\frac{dy}{dx} &=\frac{-y^2}{xy-1}\\
\frac{dy}{dx} &=\frac{y^2}{1-xy}\\
y^{'} &=\frac{y^2}{1-xy}\\
\end{align}
Clearly it is a solution of the given differential equation.
8. y -\(\cos\)y = x : \((y\sin y + \cos y +x) y^{'}=y\)
Sol: Given function is y -\(\cos\)y = x
Differentiating both sides of the equation with respect to x we get
\begin{align}
y^{'} - (-\sin y )y^{'} &= 1\\
y{'}(1+\sin y) &=1\\
y{'}&=\frac{1}{1+\sin y}\\
\end{align}
Sunbstituting this value in the given differential equation we get
\begin{aligned}
L.H.S &= (y\sin y + \cos y +x) y^{'}\\
&= (y\sin y + \cos y + y -\cos y)\frac{1}{1+\sin y}\\
&= (y\sin y + y)\frac{1}{1+\sin y}\\
&= y(1+\sin y)\frac{1}{\sin y}\\
&= y \\
&= R.H.S\\
\end{aligned}
Therefore, the given function isa solution of the given differential function.
9. x + y = \(\tan ^{-1} y \) : \(y^2y^{'} + y^2 + 1 =0\)
Sol: Given equation is
$$x + y = tan ^{-1} y $$
Differentiating both sides of the equation with respect to x we get
\begin{aligned}
1+y^{'} &= \frac{1}{1+y^2}y^{'}\\
\frac{1}{1+y^2}y^{'} - y^{'} &=1\\
y^{'}(\frac{1}{1+y^2} -1) &=1\\
y^{'}(\frac{-y}{1+y^2}&=1\\
y^{'}&=-\frac{1+y^2}{y^2}\\
\end{aligned}
Substituting the value of \(y^{'}\) in the given differential equation we get
\begin{aligned}
L.H.S &= y^2y^{'}+ y^2 + 1\\
&= y^2\times\frac{1+ y^2}{-y^2} + y^2 + 1\\
&= -1-y^2+ y^2 + 1\\
&= 0\\
&= R.H.S \\
\end{aligned}
Therefore, the given function is a solution of the given differential function.
10. \(y = \sqrt{a^2-x^2} \ x\in(-a,a)\) : \(x + y\frac{dy}{dx}=0(y\neq0)\)
Sol: Given function is
$$ y = \sqrt{a^2-x^2} \ x\in(-a,a) $$
Differentiating both sides of the equation with respect to x we get
\begin{aligned}
\frac{dy}{dx} &= \frac{-2x}{2\sqrt{a^2-x^2}}\\
&=\frac{-x}{\sqrt{a^2-x^2}}
\end{aligned}
Substituting the value of \(y^{'}\) in the given differential equation we get
\begin{aligned}
L.H.S &= x+ y\frac{dy}{dx}\\
&= x + \sqrt{a^2-x^2}\frac{x}{\sqrt{a^2-x^2}}\\
&=x-x\\
&=0\\
&= R.H.S
\end{aligned}
Therefore the given function is a solution of the given differential equation.
11. The number of arbitrary constants in the genral solution of a differential equation of fourth order are :
(a) 0 (b) 2 (c) 3 (d) 4
Sol: The correct answer is option (D)
12. The number of arbitrary constants in the particular solution of a differential equation of third order are:
(a) 3 (b) 2 (c) 1 (d) 0
Sol: The correct answer is option (D)
Class 12 Ncert Solutions Maths Differential Equations
Next Exercises
Exercise 9.1 &9.2
Exercise 9.4
Exercise 9.6
Ncert Solutions for other subjects.
Subject : General English
Question Papers
Comments
Post a Comment