{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

99_pdfsam_math 54 differential equation solutions odd

99_pdfsam_math 54 differential equation solutions odd - −...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Review Problems Next we take the partial derivative of F with respect to y and substitute 2 x 3 y for ∂F/∂y : 2 x 3 y = 2 x 3 y + g ( y ) . Thus g ( y ) = 0 and since the choice of the constant of integration is not important, we will take g ( y ) 0. Hence, we have F ( x, y ) = x 1 + x 3 y 2 and the implicit solution to the differential equation is x 1 + x 3 y 2 = C. Solving for y 2 yields y 2 = x 2 + Cx 3 . Finally we check to see if any solutions were lost in the process. We multiplied by the integrating factor µ ( x ) = x 4 so we check x 0. This is also a solution to the original equation. 21. This equation has linear coefficients. Therefore, we are looking for a substitution x = u + h and y = v + k with h and k satisfying 2 h + k 1 = 0 h + k 4 = 0 h = 1 , k = 3 . So, x = u + 1 ( dx = du ) and y = v + 3 ( dy = dv ), and the equation becomes
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( − 2 u + v ) du + ( u + v ) dv = 0 ⇒ dv du = 2 u − v u + v = 2 − ( v/u ) 1 + ( v/u ) . With z = v/u , we have v = z + uz , and so z + u dz du = 2 − z 1 + z ⇒ u dz du = 2 − z 1 + z − z = − z 2 − 2 z + 2 1 + z ⇒ z + 1 z 2 + 2 z − 2 dz = − du u ⇒ Z 1 + z z 2 + 2 z − 2 dz = − Z du u ⇒ 1 2 ln ± ± z 2 + 2 z − 2 ± ± = − ln | u | + C 1 ⇒ ² z 2 + 2 z − 2 ) u 2 = C 2 . Back substitution, z = v/u = ( y − 3) / ( x − 1), yields v 2 + 2 uv − 2 u 2 = C 2 ⇒ ( y − 3) 2 + 2( x − 1)( y − 3) − 2( x − 1) 2 = C 2 ⇒ y 2 − 8 y − 2 x 2 − 2 x + 2 xy = C. 95...
View Full Document

{[ snackBarMessage ]}