Question

Asked Sep 18, 2019

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b

Step 1

The given equation is of the form **y ^{’}+ P y = Q y^{n}** , n is not equal to

So, the given differential equation:

**y ^{’ }+ (4/x) y = x^{3}y^{2} , y(2)=-1 ......(1)**

For general solution of the given differential equation:

Multiply by **y ^{-2}** therefore the differential equation:

**y ^{-2} (y^{’ }+ (4/x) y) = y^{-2} (x^{3}y^{2}) **

** y ^{-2 }y^{’ }+ (4 /x) y^{-1 }= x^{3} ......(2)**

** **

**Let 1/y = v .......(3)**

** **

Differentiate (3) with respect to **x**, gives:

**-(1/y ^{2}) (dy /dx) =dv/dx**

** **

Equation (2) reduced in the form:

**-(dv/dx) +(4/x) v =x ^{3}**

**(dv/dx) -(4/x) v = - x ^{3}**

** **

Which is the standard form of the linear equation in **v.**

** **

Its integration factor is:

Step 2

Its solution is of the form:

Step 3

The differential equation** M(x,y) dx + N(x,y) dy...**

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