Chapter 8 Differential Equation and Applications Ex 8.1 Solutions

Balbharati Maharashtra State BoardStd 12 Commerce Statistics Part 1 Digest PdfChapter 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

(ii)
Solution:
The given D.E. is
This D.E. has highest order derivative with power 2.
∴ the given D.E. is of order 2 and degree 2.

(iii)
Solution:
The given D.E. is
This D.E. has highest order derivative 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
This D.E. has highest order derivative with power 2.
∴ the given D.E. is of order 3 and degree 2.

(v)
Solution:
The given D.E. is
On squaring both sides, we get

∴
This D.E. has highest order derivative with power 5.
∴ the given D.E. is of order 1 and degree 5.

(vi)
Solution:
The given D.E. is
This D.E. has highest order derivative with power 1.
∴ the given D.E. is of order 2 and degree 1.

(vii)
Solution:
The given D.E. is
i.e.,
This D.E. has highest order derivative with power 1.
∴ the given D.E. is of order 3 and degree 1.

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

This shows that y = xⁿis a solution of the D.E.

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

(v) y = ae^x+ be^-x
Differentiating w.r.t. x, we get
= a(e^x) + b(-e^-x) = ae^x– be^-x
Differentiating again w.r.t. x, we get
= a(e^x) – b(-e^-x)
= ae^x+ be^-x
= y
Hence, y = ae^x+ be^-xis a solution of the D.E. = y.

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

Hence, ax²+ by²= 5 is a solution of the D.E.