dy/dx+yCosx=y^nSin2x

Описание: #bernoullisequation #chinnaiahkalpana

Hello, People!

Here is the video of differential equation in bernoulli's form, which is later reduced to linear form by making simple substitution.
Observe every step to learn better.


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Chinnaiah Kalpana🍁




Note:


An equation of the form
dy/dx+Py=Qy^n
where P and Q are real numbers or functions of x alone and n s a real number such that n≠0, n≠1.
is called a Bernoulli's Equation (or) Bernoulli's Differential Equation.


An equation dx/dy+Px=Qx^n is also in Bernoulli's form. Where P and Q functions of y-alone.

If n=1, bernoulli's equation in y becomes
dy/dx+(P-Q)y=0
here the variables are separable.
Then the general solution is
∫dy/y + ∫(P-Q)dx = C.

If n=0, bernoulli's equations in y becomes
dy/dx+Py=Q(y^0)
dy/dx+Py=Q(1)
dy/dx+Py=Q which is linear.

Reducing bernoulli's to linear form:


dy/dx + Py = Qy^n -----(1)

multiplying with y^(-n);

we get,

y^(-n)dy/dx+Py^(1-n)=Q -----(2)

Let y^(1-n) = u
then (1-n)y^(-n)dy/dx = du/dx

then y^(-n)dy/dx = 1/(1-n) du/dx ----(3)

from (2) & (3)

(1/(n-1))du/dx + Pu = Q

then du/dx +(1-n)P . u = (1-n)Q ---(4)
which is linear in u and x.
Now use linear procedure to find the general solution, later replace 'u = y^(1-n)'.







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