Each of the differential equations is of two different types considered in this chapter-separable, linear, homogeneous, Bernoulli, exact, etc. Hence, derive general solutions for each of these equations in two different ways; then reconcile your results. dydx=3(y+7)x2

Name: Each of the differential equations is of two different types considered in this chapter-separable, linear, homogeneous, Bernoulli, exact, etc. Hence, derive general solutions for each of these equations in two different ways; then reconcile your results. dydx=3(y+7)x2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Each of the differential equations is of two different types considered in this chapter-separable, linear, homogeneous, Bernoulli, exact, etc. Hence, derive general solutions for each of these equations in two different ways; then reconcile your results. dydx=3(y+7)x2

Question

Each of the differential equations is of two different types considered in this chapter-separable, linear, homogeneous, Bernoulli, exact, etc. Hence, derive general solutions for each of these equations in two different ways; then reconcile your results.  dydx=3(y+7)x2

abonirali59 · Accepted Answer

Given: dydx=3(y+7)x2  ⇒dydx=3x2+21x2  Here, P=−3x2  Therefore, the integrating factor is,  I.F=e∫p dx  =e∫−3x2dx=e−x3  Therefore the solution is,  y(I.F)=∫Q(I.F)dx+c  ⇒y(e−x3)=∫21x2(e−x3)dx+c  Now to solve ∫x2e−x3dx  Plug −x3=t  ⇒−3x2dx=dt  ⇒x2dx=−dt3  ⇒∫x2e−x3dx=−13∫eudt  =∫x2e−x3dx=−13eu=−13e−x  Therefore, the solution is,  y(e−x3)=21(−13e−x3)+c  ⇒y=−7+cx3  Second method:  dydx=3(y+7)×2  dyy+7=3x2dx  Now integrate it, we get,  ln⁡(y+7)=cx3  ⇒y+cex3

autormtak0w · Accepted Answer

Seperating variables:∫dy(y+7)2⇒∫dx⋅3x2(y+7)−2+1−2+1⇒⧸3⋅x3⧸3+c−1(y+7)=x3+cThus method of seperation of variable works here.Bernauli&#039;s: y′+p(x)y=v(x)y′−3x2y=21x2dydx=f(xy)

alenahelenash · Accepted Answer

dydx=3(y+7)x2 ⇒dydx=3x2+21x2 P=−3x2 I.F=e∫p dx =e∫−3x2dx=e−x3 y(I.F)=∫Q(I.F)dx+c ⇒y(e−x3)=∫21x2(e−x3)dx+c ∫x2e−x3dx −x3=t ⇒−3x2dx=dt ⇒x2dx=−dt3 ⇒∫x2e−x3dx=−13∫eudt =∫x2e−x3dx=−13eu=−13e−x y(e−x3)=21(−13e−x3)+c ⇒y=−7+cx3 dydx=3(y+7)xx2 dyy+7=3x2dx ln⁡(y+7)=cx3 ⇒y+cex3 ⇒y=−7+cex3

Each of the differential equations is of two different types

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