Step by step solution to "Find a general solution of each of the..."

cistG

Answered question

2021-09-19

Find a general solution of each of the following equations:

x^{2} y^{'} = y^{2} + 2 x y

2abehn

Skilled2021-09-20Added 88 answers

Step 1
Given differential equation is:

x^{2} y^{'} = y^{2} + 2 x y

Rewriting given differential equation:

y^{'} - \frac{2 x}{x^{2}} y = \frac{1}{x^{2}} y^{2}

\frac{1}{y^{2}} y^{'} - \frac{2}{x y^{2}} y = \frac{1}{x^{2}}

\frac{1}{y^{2}} y^{'} - \frac{2}{x y} = \frac{1}{x^{2}}

...(1)
Let us assume

\frac{1}{y} = t

...(2)

\frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d t}{d x}

...(3)
Step 2
From equation (1), (2) and (3):

\frac{- d t}{d x} - \frac{2}{x} t = \frac{1}{x^{2}}

\frac{d t}{d x} + \frac{2}{x} t = \frac{- 1}{x^{2}}

...(4)
Linear differential equation is of the form

\frac{d y}{d x} + p y = q

O . F = e^{\int p d x}

Solution is given as:

y \times I . F . = \int q \times I . F . d x + C

Step 3
Compare equation (4) with standard linear differential equation:

p = \frac{2}{x}, q = \frac{- 1}{x^{2}}

I . F . = e^{\int \frac{2}{x} d x}

= e^{2 \log x}

= x^{2}

Solution is given as:

t x^{2} = \int \frac{- 1}{x^{2}} \times x^{2} d x + C

t x^{2} = - x + C

...(5)
Where 'C' is an integration constant.
Step 4
Replace t with

\frac{1}{y}

in equation (5)

\frac{x^{2}}{y} = - x + C

Answer:
Solution of give differential equation is:

\frac{x^{2}}{y} = - x + C

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