dy/dx+y/x=y^2xsinx
Автор: Maths Pulse - Chinnaiah Kalpana
Загружено: 2022-05-27
Просмотров: 11814
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#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.
My hearty thanks to all the subscribers, viewers, supporters ad well-wishers❤
With Love,
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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• Differential Equations- Engineering Mathem...
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