Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Differential Equation and Applications Ex 8.1

Question 1.
Determine the order and degree of each of the following differential equations:
(i) \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
Solution:
The given D.E. is \(\frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}+8=0\)
This D.E. has highest order derivative \(\frac{d^{2} x}{d t^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(ii) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=a^{x}\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=a^{x}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. is of order 2 and degree 2.

(iii) \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
Solution:
The given D.E. is \(\frac{d^{4} y}{d x^{4}}+\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
This D.E. has highest order derivative \(\frac{d^{4} y}{d x^{4}}\) with power 1.
∴ the given D.E. is of order 4 and degree 1.

(iv) (y'”)² + 2(y”)² + 6y’ + 7y = 0
Solution:
The given D.E. is (y”‘)² + 2(y”)² + 6y’ + 7y = 0
This can be written as \(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+6 \frac{d y}{d x}+7 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ the given D.E. is of order 3 and degree 2.

(v) \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
Solution:
The given D.E. is \(\sqrt{1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}}=\left(\frac{d y}{d x}\right)^{3 / 2}\)
On squaring both sides, we get
\(1+\frac{1}{\left(\frac{d y}{d x}\right)^{2}}=\left(\frac{d y}{d x}\right)^{3}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}+1=\left(\frac{d y}{d x}\right)^{5}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 5.
∴ the given D.E. is of order 1 and degree 5.

(vi) \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=7 \frac{d^{2} y}{d x^{2}}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

(vii) \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{1 / 6}=9\)
i.e., \(\frac{d^{3} y}{d x^{3}}=9^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. is of order 3 and degree 1.

Question 2.
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:

Solution:
(i) xy = log y + k
Differentiating w.r.t. x, we get

Hence, xy = log y + k is a solution of the D.E. y'(1 – xy) = y².

(ii) y = xⁿ
Differentiating twice w.r.t. x, we get

This shows that y = xⁿ is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-n x \frac{d y}{d x}+n y=0\)

(iii) y = e^x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = e^x = y
Hence, y = ex is a solution of the D.E. \(\frac{d y}{d x}\) = y.

(iv) y = 1 – log x
Differentiating w.r.t. x, we get

Hence, y = 1 – log x is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}=1\)

(v) y = ae^x + be^-x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = a(e^x) + b(-e^-x) = ae^x – be^-x
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = a(e^x) – b(-e^-x)
= ae^x + be^-x
= y
Hence, y = ae^x + be^-x is a solution of the D.E. \(\frac{d^{2} y}{d x^{2}}\) = y.

(vi) ax² + by² = 5
Differentiating w.r.t. x, we get

Hence, ax² + by² = 5 is a solution of the D.E.
\(x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}=y\left(\frac{d y}{d x}\right)\)