Question

Solve differential equation y = (1+p)x + p^2 , where p = dy/dx​

Answer 1

Answer:

Step-by-step explanation:

We have,y = (1+p)x + ap2 ....(1)differentiating with respect to x, we getdydx = 1 + p + xdpdx + 2apdpdx⇒p = 1+p+xdpdx + 2apdpdx⇒dxdp + x = −2apThe above is a linear differential equation.Now, IF = e∫dp = epNow, the solution of the differential equation is : xep = ∫−2ap×ep dp + C⇒xep = −2a[pep−ep] + C⇒xep = −2apep + 2aep + C⇒x = −2ap + 2a + Ce−p ......(2)Now, putting the value of x in (1), we gety = (1+p)(−2ap+2a+Ce−p) + ap2⇒ y = −2ap + 2a + Ce−p − 2ap2 + 2ap + pCe−p + ap2⇒ y = 2a − ap2 + (1 + p) Ce−p⇒ y = a (2 − p2) + (1 + p) Ce−p ......(3)The equation (2) and (3) constitute the required solution.

Answer 2

$\huge\red{Answer}$

We have,y = (1+p)x + ap2     ....(1)

differentiating with respect to x, we get

dy/dx = 1 + p + x dp/dx + 2ap dp/ᴅx

⇒p = 1+p+x dp/dx + 2ap dp/ᴅx

⇒dx/dp + x = −2ap

The above is a linear differential equation.

Now, IF = e∫dp = ep

Now, the solution of the differential equation is :   

 xe^p = ∫−2ap×e^p dp + ᴄ

⇒xe^p = −2a[pe^p−e^p] + ᴄ

⇒xe^p = −2ape^p + 2ae^p + ᴄ

⇒x = −2ap + 2a + Ce−p    ......(2)

Now, putting the value of x in (1), we ɢᴇᴛ

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