Respuesta :
Answer:
Therefore [tex]y=x^2+\frac{c}{x^2}[/tex] is a solution of [tex]xy'+2y=4x^2[/tex].
[tex]\therefore y=x^2-\frac{9800}{x^2}[/tex]
Step-by-step explanation:
The given differential equation is
[tex]xy'+2y=4x^2[/tex]
[tex]\Rightarrow y'+\frac{2y}{x}=4x[/tex]
Here [tex]P=\frac {2}{x}[/tex] and Q= 4x
The integrating factor is [tex]=e^{\int p dx[/tex]
[tex]=e^{\int \frac{2}{x} dx[/tex]
[tex]=e^{2\ ln x}[/tex]
[tex]= e^{lnx^2}[/tex]
[tex]=x^2[/tex]
Multiplying the integrating factor both sides of the DE
[tex]x^2y'+x^2\frac{2y}{x}=4x.x^2[/tex]
[tex]\Rightarrow x^2 \frac{dy}{dx}+2yx=4x^3[/tex]
[tex]\Rightarrow x^2dy+2yxdx=4x^3dx[/tex]
Integrating both sides
[tex]\int x^2dy+\int 2yxdx=\int 4x^3dx[/tex]
[tex]\Rightarrow x^2y= \frac{4x^4}{4}+c[/tex] [ c is an arbitrary constant]
[tex]\Rightarrow x^2y =x^4+c[/tex]
[tex]\Rightarrow y=x^2+\frac{c}{x^2}[/tex] [ dividing by x²]
Therefore [tex]y=x^2+\frac{c}{x^2}[/tex] is a solution of [tex]xy'+2y=4x^2[/tex].
The initial condition is y(10)=2
Putting x=10 and y=2
[tex]\therefore 2=10^2+\frac{c}{10^2}[/tex]
[tex]\Rightarrow 200= 10000+c[/tex] [ multiply by 100]
[tex]\Rightarrow c=200-10000[/tex]
[tex]\Rightarrow c=-9800[/tex]
Putting the value of c in the solution we get,
[tex]\therefore y=x^2-\frac{9800}{x^2}[/tex]
Yes, the function [tex]y=x^{2} +\frac{c}{x^{2} }[/tex] is solution of given differential equation.
The value of c is -9,800.
Differential equation:
The given differential equation is,
[tex]xy'+2y=4x^{2} \\\\y'+\frac{2}{x}y=4x[/tex]
Compare above equation with [tex]y'+Py=Q[/tex]
We get, [tex]P=\frac{2}{x} ,Q=4x[/tex]
Now we have to find integrating factor,
[tex]=e^{\int\limits {\frac{2}{x} } \, dx } \\\\=e^{2ln(x)}=x^{2}[/tex]
So that,
[tex]y*x^{2} =\int\limits {4x^{3} } \, dx \\\\y*x^{2} =4*\frac{x^{4} }{4} +c\\\\y=x^{2} +\frac{c}{x^{2} }[/tex]
the initial condition y(10)=2,
[tex]2=100+\frac{c}{100} \\\\c=-9800[/tex]
Hence, the value of c is -9,800.
Learn more about the differential equation here:
https://brainly.com/question/1164377
